Find the work (in joules) required to lift a 3-gram block of matter a height of 10 centimetres against the force of gravity (with g=9.8 m/sec2).
Hint+
Watch your units: 1J=1sec2kg⋅m2, but your mass is not given in kilograms, and your height is not given in metres.
Answer+
0.00294 J
Full solution+
Force is mass × acceleration (with acceleration equal to g in this problem), and both in this scenario are constant, so we don't need an integral–only a product–to calculate the force acting on the block.
To find the force in newtons, recall one newton is one sec2kg⋅m, so we need the mass of our block in kg. Specifically, our block has mass 10003 kg. So, the force involved is
F=(10003 kg)×(9.8sec2m)=0.0294sec2kg⋅m=0.0294N
To find the work in joules, recall one joule is one newton-metre: that is, one newton of force acting over one metre. So, we need our distance in metres.
A rock exerts a force of 1 N on the ground where it sits due to gravity. Use g=9.8 m/sec2.
What is the mass of the rock?
How much work (in joules) does it take to lift that rock one metre in the air?
Hint+
The force of the rock on the ground is the product of its mass and the acceleration due to gravity.
Answer+
The rock has mass 9.81 kg (about 102 grams); lifting it one metre takes 1 J of work.
Full solution+
The force of the rock is one newton, or one kilogram-metre per second squared, so
1sec2kg⋅m=(xkg)(9.8sec2m)
Therefore, the mass of the rock is 9.81 kg, or about 102 grams.
Now, since one joule is one newton-metre, the amount of work required to counteract 1 N of gravitational force for one metre is precisely one joule.
Remark: having an idea of how much work a joule is, and how much force a newton is, is a good tool for checking the reasonableness of your work. For example, after this question, if you calculate that a marble weighs 100 N, you can be pretty sure there's an error in your calculation.
where x is measured in metres and F(x) is measured in kilogram-metres per second squared (newtons).
For some large n, we might approximate
W≈i=1∑nF(xi)Δx
where Δx=nb−a and xi is some number in the interval [a+(i−1)Δx,a+iΔx]. (This is just the general form of a Riemann sum).
What are the units of Δx?
What are the units of F(xi)?
Using your answers above, what are the units of W?
Remark: we already know the units of W from the text, but the Riemann sum illustrates why they make sense arising from this particular integral.
Hint+
Adding or subtracting two quantities of the same units doesn't change the units. For example, if I have one metre of rope, and I tie on two more metres of rope, I have 1+2=3 metres of rope–not 3 centimetres of rope, or 3 kilograms of rope.
Multiplying or dividing quantities of some units gives rise to a quantity with the product or quotient of those units. For example, if I buy ten pounds of salmon for $50, the price of my salmon is 10 pounds50 dollars=1050pounddollars=5pounddollars. (Not 5 pound-dollars, or 5 pounds.)
Answer+
(a) metres (b) newtons (c) joules
Full solution+
We defined Δx=nb−a: that is, the length of one interval, when we chop [a,b] into n of them. If b and a are measured in metres, then Δx is measured in metres as well. So, the units of Δx are metres.
Put another way, since a and b both describe a quantity in metres, b−a describes a quantity in metres as well. (When we add or subtract quantities of the same units, their sum or difference is given in the same units.) Since n is a unitless quantity (simply a number: not “n kg" or “n m"), nb−a still describes a quantity in metres. (If I have 6 metres of cloth, and I cut it into 3 pieces, each piece has 36=2 metres–not 2 kilograms, or 2 metres per second.)
Since F(x) is measured in kilogram-metres per second squared (newtons), the units of F(xi) are kilogram-metres per second squared (newtons).
W is calculated by adding up summands of the form F(xi)Δx. The units of F(xi)Δx are the products of the units of F(xi) with the units of Δx. That is, the units of F(xi)Δx are (sec2kg⋅m)(m)=sec2kg⋅m2=J. The sum of terms given in joules is itself given in joules, so the units of W are joules.
Suppose f(x) has units megaFonziesmoot, and x is measured in barns (For this problem, it doesn't matter what the units measure, but a smoot is a silly measure of length; a megaFonzie is an apocryphal measure of coolness; and a barn is a humorous (but actually used) measure of area. For explanations (and entertainment) see https://en.wikipedia.org/wiki/List_of_humorous_units_of_measurement and https://en.wikipedia.org/wiki/List_of_unusual_units_of_measurement (accessed 27 July 2017).). What are the units of the quantity ∫01f(x)dx?
megaFonziesmoot⋅barn (smoot-barns per megaFonzie)
Full solution+
As we saw in Question 3, the units of ∫abf(x)dx are simply the units of the integrand, f(x), multiplied by the units of the variable of integration, x. In this case, that yields
megaFonziesmoot⋅barn (that is, smoot-barns per megaFonzie).
You want to weigh your luggage before a flight. You don't have a scale or balance, but you do have a heavy-duty spring from your local engineering-supply store. You nail it to your wall, marking where the bottom hangs. You hang a one-litre bag of water (with mass one kilogram) from the spring, and observe that the spring stretches 1 cm. Where on the wall should you mark the bottom of the spring corresponding to a hanging mass of 10kg?
You may assume that the spring obeys Hooke's law.
Hint+
Hooke's law says that the force required to stretch a spring x units past its natural length is proportional to x; that is, there is some constant k associated with the individual spring such that the force required to stretch it x m past its natural length is kx.
Answer+
10 cm below the bottom of the unloaded spring
Full solution+
Hooke's law says that the force required to stretch a spring x units past its natural length is proportional to x; that is, there is some constant k associated with the individual spring such that the force required to stretch it x m past its natural length is kx.
Since the force required to stretch the spring is proportional to the amount stretched, and the force acting on the spring is proportional to the mass hanging from it, we conclude the amount the spring stretches is proportional to the mass hung from it. So, if 1 kg stretches it 1 cm, then 10 kg will stretch it 10 cm. We should mark the wall 10 cm below the bottom of the spring as it hangs unloaded.
We can find k from the test with the bag of water. The force exerted by the bag of water was (1 kg)(9.8 m/sec2)=9.8 N=k(1 cm). So,
k=0.01 m9.8sec2kg⋅m=980sec2kg
If we hang 10 kg from the spring, gravity exerts a force of (10 kg)(9.8 m/sec2)=98sec2kg⋅m. This will be matched by the spring with a force of kx newtons, where k is the spring constant and x is the amount stretched.
The work done by a force in moving an object from position x=1 to x=b is
W(b)=−b3+6b2−9b+4
for any b in [1,3]. At what position x in [1,3] is the force the strongest?
Hint+
Definition 2.1.1 in the CLP-2 text tells us the work done by the force from x=1 to x=b is
W(b)=∫1bF(x)dx, where F(x) is the force on the object at position x. To recover the equation for F(x), use the Fundamental Theorem of Calculus.
Answer+
x=2
Full solution+
Definition 2.1.1 in the CLP-2 text tells us the work done by the force is
W(b)=∫1bF(x)dx, where F(x) is the force on the object at position x.
So, by the Fundamental Theorem of Calculus Part 1,
A variable force F(x)=xa Newtons moves an object
along a straight line when it is a distance of x meters from the origin.
If the work done in moving the object from x=1 meters to
x=16 meters is 18 joules, what is the value of a? Don't worry
about the units of a.
Hint+
Review Definition 2.1.1 in the
CLP-2 text for calculating the work done by a force over a distance.
Answer+
a=3
Full solution+
By Definition 2.1.1 in the CLP-2 text, the work done in moving the object from x=1 meters
to x=16 meters by the force F(x) is
W=∫116F(x)dx=∫116xadx=[2ax]x=1x=16=6a
To have W=18, we need a=3.
As a side remark, F(x)=xa should have units Newtons.
Since x, a distance, is measured in meters, a has to have the bizarre
units newton-meters.
A tube of air is fitted with a plunger that compresses the air as it is pushed in. If the natural length of the tube of air is ℓ, when the plunger has been pushed x metres past its natural position, the force exerted by the air is ℓ−xc N, where c is a positive constant (depending on the particulars of the tube of air) and x<ℓ.
What are the units of c?
How much work does it take to push the plunger from 1 metre past its natural position to 1.5 metres past its natural position? (You may assume ℓ>1.5.)
Hint+
For (a), ℓ−xc is meausured in Newtons, while ℓ and x are in metres. For (b), notice the similarities and differences between the tube of air and a spring obeying Hooke's law.
Answer+
(a) joules (b) clog(ℓ−1.5ℓ−1) J
Full solution+
Since ℓ−xc is measured in newtons, and ℓ and x (and therefore ℓ−x) are measured in metres, the units of c are newton-metres, i.e. joules.
Following Definition 2.1.1 in the CLP-2 text, the work done compressing the air is
W=∫11.5F(x)dx
where F(x) is the amount of force applied when the plunger is x metres past its natural position. The amount of force applied is equal in magnitude to the amount of force supplied by the tube: ℓ−xc N. Note ℓ and c are constants. We can guess the antiderivative, or use the substitution u=ℓ−x, du=−dx.
Note that, because ℓ>1.5, the argument of logarithm is positive, so we don't need the absolute value signs. Furthermore, ℓ−1>ℓ−1.5, so ℓ−1.5ℓ−1>1, hence log(ℓ−1.5ℓ−1)>0.
Questions 9 through 16 offer practice on two broad types of calculations covered in the text: lifting things against gravity, and stretching springs. You may make the same physical assumptions as in the text: that is, springs follow Hooke's law, and the acceleration due to gravity is a constant −9.8 metres per second squared.
Find the work (in joules) required to stretch a spring 10 cm beyond
equilibrium, if its spring constant is k=50N/m.
Hint+
See Example 2.1.2 in the
CLP-2 text. Be careful about your units.
Answer+
41J
Full solution+
By Hooke's Law, the force exerted by the spring
at displacement x m from its natural length is F=kx,
where k is the spring constant.
Measuring distance in meters and force in newtons (since one joule is one newton-metre), the total work is
Note the units of the integrand (kx) are newtons, and the units of the variable of integration, x, are metres. So, the evaluated integral has units newton-metres, or joules.
A force of 10 N (newtons) is required to hold
a spring stretched 5 cm beyond its natural length. How much work,
in joules (J), is done in stretching the spring from its
natural length to 50 cm beyond its natural length?
Hint+
Be careful about the units.
Answer+
25 J
Full solution+
First note that newtons and joules are SI units with one joule equal to one newton-metre, so we should
measure distances in meters rather than centimeters.
Next recall that a spring with spring constant k
exerts a force F(x)=kx when the spring is stretched x m beyond its
natural length. So in this case (0.05 m)(k)=10 N, or k=200 N/m. The work done is:
∫00.5mF(x)dx=∫00.5200xdx=[100x2]00.5=25J
Note the units of the integrand (F(x)=kx=200x) are newtons (k is given in N/m, and x is given in m). The units of the variable of integration, x are metres. So, the evaluated integral has units newton-metres, or joules.
A 5-metre-long cable of mass 8 kg is used to lift
a bucket off the ground. How much work is needed
to raise the entire cable to height 5 m? Ignore the mass of the bucket and its contents.
Hint+
Suppose that the bucket is a distance y above the ground.
How much work is required to raise it an additional height dy?
Answer+
196J
Full solution+
Note that the cable has mass density 58 kg/m.
When the bucket is at height y, the cable that remains to be lifted has length (5−y) m
and mass 58(5−y)=8(1−5y) kg.
So, at height y, the cable is subject to a downward gravitational force
of 8(1−5y)⋅9.8 N; to raise the cable we need
to apply a compensating upward force of 8(1−5y)⋅9.8 N.
So, the work required is
Alternatively, the cable has linear density 8 kg/5 m=1.6 kg/m, and so the work required to lift a small piece of the cable (of length Δy) from height y m to height 5 m is
A tank 1 metre high has pentagonal cross sections of area 3 m2 and is filled with water. How much work does it take to pump out all the water?
You may assume the density of water is 1 kg per 1000 cm3.
Hint+
Since you're given the area of the cross-section, it doesn't matter what shape it has. However, the density of water is given in cubic centimetres, while the measurements of the tank are given in metres.
Answer+
14700 J
Full solution+
Imagine pumping out a thin, horizontal layer of water that is at height y–that is, y metres above the bottom of the tank. Let the width of the layer be dy.
The volume of water in the layer is 3dy m3 (since the cross-section has area 3 m3).
One cubic metre is equal to 1003 cubic centimetres. So, the mass of water in one cubic metre is 10001003=1000 kg.
Therefore, the mass of water in our layer is (3000dy) kg.
The force of gravity acting on it is (−9.8×3000dy) N, so we need to pump with a compensating force of (9.8×3000dy) N.
The water needs to be pumped a distance of 1−y metres.
So, the work required to pump out the thin layer of water at height y is (9.8×3000×(1−y)dy) J.
So, all together, the work to pump out the entire tank is
A sculpture, shaped like a pyramid 3m high sitting on the ground, has been made by
stacking smaller and smaller (very thin) iron plates on top of one another. The iron plate
at height z m above ground level is a square whose side length is (3−z) m. All of
the iron plates started on the floor of a basement 2 m below ground level.
Write down an integral that represents the work, in joules, it took to move all of the
iron from its starting position to its present position. Do not evaluate the
integral. (You can use 9.8 m/s2 for the acceleration due to gravity and
8000 kg/m3 for the density of iron.)
Hint+
Consider the work done to lift a horizontal plate from 2 m below the ground to a height z. You'll need to know the mass of the plate, which you can calculate from its volume, since its density is given to you.
Answer+
∫03(9.8)(8000)(2+z)(3−z)2dz joules
Full solution+
We can model the sculpture as a collection of thin horizontal plates of width dz. Remember work is force times distance; a horizontal plate at height z moved z+2 metres from the basement to its final position. So, we need to know the force acting on the plate, which is the product of the mass of the plate with the acceleration due to gravity. Since we are given the density of iron, if we find the volume of the plate, then we can calculate its mass.
The plate at height z
has side length 3−z m and hence
has area (3−z)2 m2 and hence
has volume (3−z)2dz m3 and hence
has mass 8000(3−z)2dz kg and hence
is subject to a gravitational force of
9.8×8000(3−z)2dz N and hence
requires work 9.8×8000(2+z)(3−z)2dz J
to raise it from 2 m below ground level to z m above ground
level.
Suppose a spring extends 5 cm past its natural length when one kilogram is hung from its end. How much work is done to extend the spring from 5 cm past its natural length to 7 cm past its natural length?
Hint+
You can find the spring constant k from the information about the hanging kilogram.
Answer+
0.2352 J
Full solution+
From the information given about the hanging kilogram, we can find the spring constant k. One kilogram generates a force of 9.8 N under gravity. (We find this by the calculation (1 kg)×(9.8m/sec2)=9.8 N.) This force is matched by the force of the spring, which by Hooke's law is equal to k(201m). So,
k=201 m9.8 N=196mN
Again by Hooke's law, the force required to stretch the spring x metres past its natural length is 196x N (when x is measured in metres).
So, the work required to stretch the spring from 5 cm past its natural length to 7 cm past its natural length is
Ten kilograms of firewood are hoisted on a rope up a height of 4 metres to a second-floor deck. If
the total work done is 400 joules, what is the mass of the 4 metres of rope?
You may assume that the rope has the same density all the way along.
Hint+
Follow the method of Example 2.1.6 in the CLP-2 text and Question 11 in this section.
Answer+
4920 kg, or about 408 grams
Full solution+
Let M be the mass of the rope. Then its density is 4M kg/m. Following the method of Example 2.1.6 in the CLP-2 text, we let y be the height of the firewood above the ground, so the wood is raised from y=0 to y=4. When the wood is at height y,
the rope that remains to be lifted has length 4−y, and so it has mass 4M(4−y) kg,
and the firewood still has mass 10 kg.
The remaining rope and the wood are subject to a downward gravitational force of magnitude mass[4M(4−y)+10]×9.8 N.
So, to raise the firewood from height y to height (y+dy), we need to apply a compensating upward force of [4M(4−y)+10]×9.8 through distance dy. This takes work [4M(4−y)+10]×9.8dy J.
All together, the work involved in hauling up the wood is
Since the work was 400 joules, solving 400=9.8(2M+40) for M tells us the mass of the rope is 9.8200−20=4920 kg, or about 408 g.
Alternately, the work involved in lifting up the wood is 10×9.8×4=392 J, so the work in lifting up the rope is 8 J. A small section of rope of length dy, that starts at height y above the ground, has mass 4Mdy kg and is lifted (4−y) metres, so the work involved in lifting this section of rope is 9.8×(4−y)×4Mdy. Then the amount of work to lift the whole rope (but not the wood) is
A 5 kg weight is attached to the middle of a 10-metre long rope, which dangles out a window. The rope alone has mass 1 kg. How much work does it take to pull the entire rope in through the window, together with the weight?
Hint+
Calculating the work done on the rope and the weight separately makes the computation somewhat easier.
Answer+
294 J
Full solution+
For Questions 11 and 15 in this section, we gave two methods for finding the work involved in pulling up a cable: one where we consider pulling up the entire remaining cable a tiny distance of dy, and one where we consider pulling a tiny slice of cable of length dy the entire distance up.
There is another variation we can consider with the weight: we can either calculate the work done on the weight and the work done on the rope separately, or we can calculate them together. If we calculate them together, then there are two cases to consider: the work done pulling up the first 5 metres of rope involves the weight, while the last 5 metres does not. These two choices (how to model the rope, and how to deal with the weight) actually lead to four solutions, but to avoid unnecessary repetition only two are presented below.
In this solution, we consider the work on the rope separately from work on the weight, and we imagine lifting a tiny piece of rope the entire distance to the window.
The weight has a mass of 5 kg, and is lifted a distance of 5 m to the window. The force of gravity acting on the weight is (5 kg)(9.8 m/sec2)=49 N, so the work to lift it 5 metres is (49 N)(5 m)=245 J.
The density of the rope is 101 kg/m. A tiny piece of rope of length dy, hanging y metres from the window, has mass (101dy) kg, and needs to be lifted y metres. So, the force of gravity acting on the piece of rope is (101dy kg)(9.8 m/sec2)=0.98dy N, and the work to pull it up to the window is (0.98ydy) J. So, the total work to pull up the rope is
∫0100.98ydy=0.98[2y2]010=49 J
All together, the work to pull up the rope with the weight is 245+49=294 J.
In this solution, we consider the work on the rope together with the weight, and we imagine lifting the remaining rope a tiny distance to the window.
Suppose y metres of the rope have been pulled in, and 0≤y≤5 (shown on the left, below). Then the remaining rope has length 10−y, and contains the weight, so the mass remaining to be pulled up is rope101(10−y)+weight1015=6−10y kg. Then the force of gravity acting on the dangling rope and weight is (9.8 m/sec2)((6−10y) kg)=(58.8−0.98y) N. The work needed to lift this rope dy metres is (58.8−0.98y)dy J.
Now, suppose y metres of the rope have been pulled in, and 5<y≤10 (shown above, right). Then the remaining rope has length 10−y, but does not contain the weight, so the mass remaining to be pulled up is 101(10−y)=1−10y kg. Then the force of gravity acting on the dangling rope is (9.8 m/sec2)((1−10y) kg)=(9.8−0.98y) N. The work needed to lift this rope dy metres is (9.8−0.98y)dy J.
A box is dragged along the floor. Friction exerts a force in the opposite direction of motion from the box, and that force is equal to μ×m×g, where μ is a constant, m is the mass of the box and g is the acceleration due to gravity. You may assume g=9.8 m/sec2.
How much work is done dragging a box of mass 10 kg along the floor for three metres if μ=0.4?
Suppose the box contains a volatile substance that rapidly evaporates. You pull the box at a constant rate of 1 m/sec for three seconds, and the mass of the box at t seconds (0≤t≤3) is (10−t) kilograms. If μ=0.4, how much work is done pulling the box for three seconds?
Hint+
When you pull the box, the force you're exerting is exactly the same as the frictional force, but in the opposite direction. In (a), that force is constant. In (b), it changes. Check Definition 2.1.1 in the CLP-2 text for how to turn force into work.
Answer+
(a) 117.6 J (b) 3.92[30−23]≈104 J
Full solution+
The frictional force is μ×m×g=0.4(10 kg)(9.8sec2m)=39.2sec2kg⋅m=39.2 N. Since this constant force acts over a distance of 3 metres, the work is 3×39.2=117.6 J.
In the case of a constant force, we don't need to use an integral, but we could if we wanted:
W=∫0339.2dx=[39.2x]03=39.2×3=117.6 J.
Since the box is moving at a speed of 1 m/sec, at time t we can say the box is at position t, 0≤t≤3. At position t, the mass of the box is (10−t) kg, so the frictional force is 0.4×m×g=0.4(10−tkg)(9.8sec2m)=3.92(10−t)N. Now that we know the force, to find the work we simply integrate, following Definition 2.1.1 in the CLP-2 text:
For Questions 18 and 19,
use the principle (introduced after Definition 2.1.1 in the CLP-2 text and utilized in Example 2.1.5) that the work done on a particle by a force over a distance is equal to the change in kinetic energy of that particle.
A ball of mass 1 kg is attached to a spring, and the spring is attached to a table. The ball moves with some initial velocity, and the spring slows it down. At its farthest, the spring stretches 10 cm past its natural length. If the spring constant is 5 N/m, what was the initial velocity of the ball?
You may assume that the ball starts moving with initial velocity v0, and that the only force slowing it down is the spring. You may also assume that the spring started out at its natural length, it follows Hooke's law, and when it is stretched its farthest, the velocity of the ball is 0 m/sec.
Hint+
Remember that the work done on an object is equal to the change in its kinetic energy, which is 21mv2, where m is the mass of the object and v is its velocity. Hooke's law will tell you how much work was done stretching the spring.
Answer+
251 m/sec, or about 22.36 cm/sec
Full solution+
Definition 2.1.1 in the CLP-2 text is justified by showing that the work done by a force acting on a particle is equal to the change in the kinetic energy of that particle. We can use Hooke's law to calculate the work done stretching the spring. That work will be equal to the change in kinetic energy of the ball.
The ball initially has kinetic energy 21(1 kg)(v0 m/sec)2=2v02sec2kg⋅m2=2v02 J. At the time the spring is stretched its farthest, the ball's velocity is 0 m/sec, so its kinetic energy is 21(1 kg)(0 m/sec)2=0J. So, the change in kinetic energy of the ball is 2v02 J.
Now let's find the work done by the spring. Its spring constant is k=5 N/m, so, the force on the spring when it is stretched x metres past its natural length is 5x N. The spring is stretched from its natural length to 10 cm, which is 0.1 m. Then the work done by the spring is
A mild-mannered university professor who is definitely not a spy notices that when their car is on the ground, it is 2 cm shorter than when it is on a jack. (That is: when the car is on a jack, its struts are at their natural length; when on the ground, the weight of the car causes the struts to compress 2 cm.) The university professor calculates that if they were to jump a local neighborhood drawbridge, their car would fall to the ground with a speed of 4 m/sec. If the car can sag 20 cm before important parts scrape the ground, and the car has mass 2000 kg unoccupied (2100 kg with the professor inside), can the professor, who is certainly not involved in international intrigue, safely jump the bridge?
Assume the car falls vertically, the struts obey Hooke's law, and the work done by the struts is equal to the change in kinetic energy of the car + professor. Use 9.8 m/sec2 for the acceleration due to gravity.
Hint+
As in Question 18 in this section, the change in kinetic energy of the car is equal to the work done by the compressing struts. The only added step is to calculate the spring constant, given that a car with mass 2000 kg compresses the spring 2 cm in Earth's gravity. You're not calculating work to find the spring constant: you're using the fact that when the car is sitting still, the force exerted upward by the struts is equal to the force exerted downward by the mass of the car under gravity.
Answer+
yes (at least, the car won't scrape the ground)
Full solution+
The setup to answer this question is similar to Question 18 in this section: the work done by a spring on the occupied vehicle will be equal to the change in kinetic energy of that occupied vehicle. So, we need to find the work done by the spring, and the kinetic energy lost by the falling car. In order to find the work done by the spring, we need to find the spring constant.
The car's mass of 2000 kg compresses the struts 2 cm past their natural length. The force of the car under gravity is (2000 kg)×(9.8 m/sec2)=19600 N. This force is exactly the same as that exerted by the spring, k(0.02 m). So, k=980000 N/m.
The spring can safely compress 20 cm. So, the amount of work done by the spring compressing that far gives us the maximum amount of work the spring can safely do. While the car is falling, the spring is at its natural length, so the work done to compress it to 20 cm (0.2 m) shorter is:
When the car first hits the pavement, it's falling at 4 m/sec, so it has kinetic energy 21(2100 kg)(4 m/sec)2=16800 J. When the car compresses the springs as far as they go and it starts to rebound, it has kinetic energy 0, since its instantaneous velocity is zero. So, the change in kinetic energy is
16800 J.
Since the change in kinetic energy is 16800 J, and the struts can safely do a work of (up to) 19600 J, the jump is within the (meagre) safety limits set by the question.
3Stage 3Application
Further than practice: several ideas at once, or an unfamiliar situation.
A disposable paper cup has the shape of a right circular cone with radius 5 cm and height 15 cm, and is completely filled with water. How much work is done sucking all the water out of the cone with a straw?
You may assume that 1 m3 of water has mass 1000 kilograms, the acceleration due to gravity is −9.8 m/sec2, and that the water moves as high up as the very top of the cup and no higher.
Hint+
To find the radius of a horizontal layer of water, use similar triangles. Be careful with centimetres versus metres.
Answer+
≈0.144 J
Full solution+
Let's consider sucking up a flat, horizontal layer of water. If the water is y metres above bottom of the cone, then it needs to be raised 0.15−y metres. So, if its mass is m kg, then the force of gravity acting on it is 9.8m N and the work involved in slurping it to the top of the cone is 9.8m(0.15−y) J. So, what we need to find is the mass of a layer of water y metres from the bottom of the cone.
A horizontal cross-section of the cone is a circle. To find its radius, we use similar triangles: yr=0.150.05, so r=31y. Therefore, the area of the cross-section of the cone y metres above its bottom is π(31y)2=9πy2 m2. If this layer has height dy, then its volume is
9πy2dy m3, and its mass is 10009πy2dy kg.
Now, we know that the work to suck up the layer of water y metres from the bottom of the cup is 9.8(0.15−y)(10009πy2dy) J. So, the work involved in drinking all the water is:
A spherical tank of radius 3 metres is half–full of water.
It has a spout of length 1 metre sticking up from the top of
the tank. Find the work required to pump all of the water
in the tank out the spout. The density of water is 1000 kilograms
per cubic metre. The acceleration due to gravity is 9.8 metres
per second squared.
Hint+
See Example 2.1.4 in the
CLP-2 text for a basic method for calculating the work done pumping water.
To find the area of a horizontal layer of water, use some geometry. A horizontal cross-section of a sphere is a circle, and its radius will depend on the height of the layer in the tank.
Answer+
904,050πJ
Full solution+
Imagine slicing the water into horizontal pancakes of thickness dx as in the sketch below.
Denote by x the distance of a pancake below the surface of
the water. (So, x runs from 0 to 3.) Each pancake:
has radius 32−x2 m (by Pythagoras) and hence
has cross–sectional area π(9−x2) m2 and hence
has volume π(9−x2)dx m3 and hence
has mass 1000π(9−x2)dx kg and hence
is subject to a gravitational force of
9.8×1000π(9−x2)dx N and hence
requires work 9800π(9−x2)(x+4)dx J
to raise it to the spout.
(It has to be raised x m to bring it to the height of the centre
of the sphere, then 3 m more to bring it to the top of the sphere,
and finally 1 m more to bring it to the spout.)
A 5-metre cable is pulled out of a deep hole, where it was dangling straight down. The cable has density ρ(x)=(10−x) kg/m, where x is the distance from the bottom end of the rope. (So, the bottom of the cable is denser than the top.) How much work is done pulling the cable out of the hole?
Hint+
The basic ideas you've used already with “cable problems" still work, you only need to take care that the density of the cable is no longer constant. The mass of a tiny piece of cable, say of length dx, is (density)×(length) = (10−x)dx, where x is the distance of our piece from the bottom of the cable.
If you want more work to reference, Question 22 in Section 1.6 finds the mass of an object of variable density.
Answer+
102065 J
Full solution+
Let's consider the work involved in lifting up a small section of cable, with length dy, distance y from the bottom end of the cable.
The distance this section must travel is (5−y) metres, so if its mass is M(y), then the work involved is
W=∫059.8×(5−y)×M(y)
So, we need to find M(y). The length of the section of cable is dy, and its distance from the end of the cable is y, so the mass of the section is (10−y)dy. Therefore,
Alternately, we can continue to use the basic method of Example 2.1.6 in the CLP-2 text, noticing that the density of the cable is no longer constant.
Let's consider pulling the cable up a tiny distance of dy metres, after we have already lifted it y metres (so (5−y) metres of the cable is still in the hole).
If R(y) is the mass of the remaining cable (in kg), then the force of gravity is −9.8×R(y), so the work done is 9.8×R(y)×dy. Once we find R(y), we can calculate the total work done:
W=∫059.8×R(y)×dy(∗)
As given in the question statement, the density of the cable is (10−x) kg/m, where x is the distance from the bottom end of the cable. Consider a tiny section of cable x metres from the bottom end, of length dx.
The mass of this tiny section is (10−xmkg)×(dxm)=(10−x)dx kg. The section of cable dangling is the last (5−y) metres of cable. So, the combined mass of the section of cable dangling, after we've already pulled up y metres of it, is
A rectangular tank is fitted with a plunger that can raise and lower the water level by decreasing and increasing the length of its base, as in the diagrams below. The tank has base width 1 m (which does not change) and contains 3 m3 of water.
The force of the water acting on any tiny piece of the plunger is PA, where P is the pressure of the water, and dA is the area of the tiny piece. The pressure varies with the depth of the piece (below the surface of the water). Specifically, P=cD, where D is the depth of the tiny piece and c is a constant, in this case
c=9800 N/m3.
If the length of the base is 3 m, give the force of the water on the entire plunger. (You can do this with an integral: it's the sum of the force on all the tiny pieces of the plunger.)
If the length of the base is x m, give the force of the water on the entire plunger.
Give the work required to move the plunger in so that the base length changes from 3 m to 1 m.
Hint+
To calculate the force on the entire plunger, first find the force on a horizontal rectangle with height dy at depth y.
Checking units can be a good way to make sure your calculation makes sense.
Answer+
(a) 4900 N (b) x244100 N (c) 29400 J
Full solution+
The force of the depends on depth, which varies. So, consider a thin rectangle of the plunger at depth y, with height dy and width 1 m (the width of the entire plunger). Let the area of this rectangle be dA.
The area of this rectangle is 1dy m2, so the force of the water acting on it is F=P⋅dA=c(9800m3N)d(y m)dA(dy m2)=9800ydy N.
The depth at the top of the plunger is y=0. To find the depth at the bottom of the plunger, note that the water has a volume of 3 m3, and is in a rectangular container with base 1 m by 3 m. So, its height is 1 m.
The force over the entire plunger, from depth y=0 to y=1, is
∫019800ydy=[4900y2]01=4900 N
Let's follow our work from part (a), but with the width of the length of the base as x m.
Still, a thin rectangle of plunger has width 1 m and height dy m, so it has area dy m2. At depth y, it has a force from the water of 9800ydy N. This hasn't changed from (a).
Now, let's consider the depth of the water. The volume of water is 3 m3, and it is in a rectangular container with base 1 m by x m. So, its depth is 3/x m.
Therefore, the force on the entire plunger must be calculated from y=0 to y=3/x.
F(x)=∫03/x9800ydy=[4900y2]03/x=x294900=x244100 N
Let's check that this answer makes sense: F(3)=4900 N, which matches our answer from (a).
If the force of water acting on the plunger, when the length of the base is x metres, is given by F(x), then we push the plunger with a force of −F(x). Then the work we're looking for is
W=∫31−F(x)dx=∫13F(x)dx
F(x) is exactly what we found in (b):
F(x)=x244100 N.
A leaky bucket picks up 5 L of water from a well, but drips out 1 L every ten seconds. If the bucket was hauled up 5 metres at a constant speed of 1 metre every two seconds, how much work was done?
Assume the rope and bucket have negligible mass and one litre of water has 1 kg mass, and use 9.8 m/sec2 for the acceleration due to gravity.
Hint+
When y metres of rope have been hauled up, what is the mass of the water?
Answer+
220.5 J
Full solution+
Let's start by converting from time spent pulling to amount pulled. When y metres of rope have been pulled up, 2y seconds have passed, so 51y litres of water have leaked out of the bucket, leaving 5−51y litres. (This only makes sense when 51y≤5, but we only consider values of y from 0 to 5, so it's not a problem. That is, we're never hauling up an empty bucket that can't leak any more.)
When we've pulled up y metres of rope, the mass in the bucket is (5−51y) kg, so the force of gravity acting on it is 9.8(5−51y) N. Since we pull up 5 metres of rope, the work done is:
The force of gravity between two objects, one of mass m1 and another of mass m2, is F=Gr2m1m2, where r is the distance between them and G is the gravitational constant.
How much work is required to separate the earth and the moon far enough apart that the gravitational attraction between them is negligible?
Assume the mass of the earth is 6×1024 kg
and the mass of the moon is 7×1022 kg, and that they are currently 400000 km
away from each other. Also, assume G=6.7×10−11kg⋅sec2m3, and the only force acting on the earth and moon is the gravity between them.
Hint+
The work you're asked for is an improper integral, moving the earth and moon infinitely far apart.
Answer+
About 7×1028 J
Full solution+
According to the formula for gravity between two objects, the earth and moon will gravitationally attract one another no matter how far apart they are, so what we're looking for is the work to separate them infinitely far. That is, we want to calculate
∫a∞F(r)dr, where a=400000000 m.
If we take m1 and m2 to be the mass of the earth and moon as given in the question statement, then:
Remark: since the force of gravity between the earth and the moon gets weaker as they are farther apart, it takes less and less work to move them each kilometre. If we move them a finite distance apart, the work involved will always be less than 7×1028 joules, no matter how huge that finite distance is. If we move them a very, very long (but finite) distance apart, the work we did will be quite close to (but still less than) 7×1028 joules.
True or false: the work done pulling up a dangling cable of length ℓ and mass m (with uniform density) is the same as the work done lifting up a ball of mass m a height of ℓ/2.
Hint+
You can formulate a guess by considering the work done on the ball versus the work done on the rope in Question 16, Solution 1. But be careful–the ball in that problem did not have the same mass as the rope.
Answer+
true
Full solution+
A ball of mass m experiences a gravitational force of mg, so lifting it a height of ℓ/2 involves a work of 21mgℓ.
The cable has density m/ℓ. A tiny section of cable with length dy has mass ℓmdy, and so gravity acts on it with a force of ℓmgdy. If the tiny section of cable is y units from the top of the cable, it needs to be pulled up y units, so the work on that section is ℓmgydy. Therefore, the work to pull up the entire cable is
∫0ℓℓmgydy=[2ℓmgy2]0ℓ=2ℓmgℓ2=21mgℓ
So, the work to pull up a cable with uniform density is the same as the work to pull up a ball with the same mass from the middle height of the cable.
Remark: this is a nice fact to use when you're checking your computations for “pulling up cable" problems, but keep in mind it depends on the cable being of uniform density.
A tank one metre high is filled with watery mud that has settled to be denser at the bottom than at the top.
At height h metres above the bottom of the tank, the cross-section of the tank has the shape of the finite region bounded by the two curves y=x2 and y=2−h−3x2.
At height h metres above the bottom of the tank, the density of the liquid is 10002−h kilograms per cubic metre.
How much work is done to pump all the liquid out of the tank?
You may assume the acceleration due to gravity is 9.8 m/sec2.
Hint+
There are two things that vary with height: the density of the liquid, and the area of the cross-section of the tank. Make a formula M(h) for the mass of a thin layer of liquid h metres below the top of the tank, using mass=volume×density. The rest of the problem is similar to other tank-pumping problems in this section.
Answer+
925595 J
Full solution+
Like our other tank-pumping problems (e.g. Questions 12, 20, and 21 in this section, and Example 2.1.4 in the CLP-2 text), we can find the work done by considering thin layers of liquid. If the layer of liquid h metres above the bottom of the tank with thickness dh has mass M(h), then the force of gravity acting on it is −9.8M(h) N and the work required to pump it to the top of the tank (1−h metres away) is 9.8(1−h)M(h) J. So, the work to empty the entire tank is
W=∫019.8(1−h)M(h)(∗)
Our remaining task is to find M(h). There are two things that vary with height: the density of the liquid, and the area of the cross-section of the tank.
At height h metres, the cross-section of the tank is shaped like the finite region bounded by the curves y=x2 and y=2−h−3x2. To find this area, we need an integral (see Section 1.5 for a refresher), and to find the limits of integration, we need to know where the two curves meet. By solving x2=2−h−3x2, we find that they meet at x=±212−h. (Recall h is between 0 and 1, so 2−h is a real number, i.e. the curves do indeed meet.) Furthermore, when −212−h≤x≤212−h, then x2≤2−h−3x2, so y=x2 is the bottom function and 2−h−3x2 is the top function.
So, (taking advantage of the fact that our region has even symmetry) the area of the cross-section of the tank at height h is
An hourglass is 0.2 m tall and shaped such that that y metres above or below its vertical centre it has a radius of y2+0.01 m.
It is exactly half-full of sand, which has mass
M=71 kilograms.
How much work is done on the sand by quickly flipping the hourglass over?
Assume that the work done is only moving against gravity, with g=9.8 m/sec2, and the sand has uniform density. Also assume that at the instant the hourglass is flipped over, the sand has not yet begun to fall, as in the picture above.
Hint+
You can model the motion, instead of a rotation, as dividing the sand into thin horizontal slices and lifting each of them to their new position.
In order to calculate the work involved lifting a layer of sand, you need to know the mass of the layer of sand.
To find the mass of a layer of sand, you need its volume and the density of the sand.
To find the density of the sand, you need to the volume of the sand: that is, the volume of half the hourglass.
The hourglass is a solid of rotation: you can find its volume using an integral, as in Section 1.6.
Answer+
407=0.175 J
Full solution+
Since the only work done is against the force of gravity, we only need to know how high the sand was lifted, not how it got there. So, we don't really need to worry about its semicircular path: we can imagine that every grain of sand was lifted from its old position to its new position.
Consider a thin, horizontal layer of sand in the hourglass, y metres below the vertical centre of the hourglass.
Its final position is y metres above the centre of the hourglass. That is, it was lifted 2y metres against the force of gravity.
The layer is shaped like a circle with radius y2+0.01 and height dy, so its volume is π(y2+0.01)2dy cubic metres.
To find the mass of the layer, we need to know the density of the sand. Let the volume of sand in the hourglass be V. We are given its mass M. Then π(y2+0.01)2dy cubic metres has a mass of VMπ(y2+0.01)2dy kilograms.
So, the force of gravity acting on the layer is 9.8VMπ(y2+0.01)2dy N, acting vertically downwards.
To lift the layer to its final position, we apply a compensating force over a distance of 2y metres, for a total work of 9.8VMπ(y2+0.01)22ydy J.
Since the hourglass has height 0.2 m, and exactly half of it is filled with sand, the top layer of sand is exactly at the vertical centre of the hourglass, and the bottom layer of sand is 0.1 metres below.
Using V is the volume of sand in the hourglass, and M is its mass (we're given M=71 kg) then the total work flipping all the sand is:
It remains to find V: the volume of sand in the hourglass. We know the sand is in the shape of a solid of rotation. Recall from Section 1.6 that we can find the volume of such shapes by slicing them into thin disks.
In the picture above, we've used an axis that matches the way we've been describing our solid: 0 is the vertical centre of the hourglass, which is where the top of the sand is, and the bottom of the sand is 0.1 metres from 0.
To find the volume of this solid, we slice it into horizontal disks. The disk that is x metres from the centre of the hourglass has radius x2+0.01 and thickness dx, so it has volume π(x2+0.01)2dx. The volume of the entire solid, i.e. the volume of the sand, is:
Suppose at position x a particle experiences a force of F(x)=1−x4 N. Approximate the work done moving the particle from x=0 to x=1/2, accurate to within 0.01 J.
Hint+
Theorem 1.11.12 in the CLP-2 text gives error bounds for the standard types of numerical approximations. You won't need very many intervals to achieve the desired accuracy.
Answer+
One possible answer: 411−(81)4+1−(83)4
Full solution+
Using Definition 2.1.1 in the CLP-2 text, the work involved is
W=∫01/21−x4dx J.
However, the function F(x)=1−x4 happens to not have an antiderivative that can be expressed as an elementary function. That means we can't use the Fundamental Theorem of Calculus Part 2 to evaluate this integral (at least, not without knowing a bit more about functions than is prerequisite for this course). Instead, we can use numerical methods, like the midpoint rule or Simpson's rule, to approximate its value.
It's not immediately clear which rule (Simpson's, midpoint, or trapezoidal) will lead us down the easiest path. For Simpson's rule, we need to know the fourth derivative of F(x), which is not a simple task. But, we often need fewer intervals for Simpson's rule than for the midpoint or trapezoid rules. In this case, we'll show below that n=2 intervals suffice to guarantee a low enough error using the midpoint rule, so Simpson's rule won't let us get away with fewer intervals. Below, we find the approximation using the midpoint rule–but there are other ways as well.
In order to decide how many intervals we should use with the midpoint rule, we need to know the second derivative of F(x).
