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Applications of Integration

2.1 Work

29 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

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=9.8g=9.8 m/sec2^2).

Hint

Watch your units: 1 J=1 kg⋅m2sec21 \,\mathrm{J} = 1\, \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{sec}^2}, but your mass is not given in kilograms, and your height is not given in metres.

Answer

0.00294 J

Full solution
Q2Stage 1

A rock exerts a force of 1 N on the ground where it sits due to gravity. Use g=9.8g=9.8 m/sec2^2.

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 19.8\dfrac{1}{9.8} kg (about 102 grams); lifting it one metre takes 1 J of work.

Full solution
Q3Stage 1

Consider the equation

W=∫abF(x) dxW = \int_a^b F(x)\,\dee{x}

where xx is measured in metres and F(x)F(x) is measured in kilogram-metres per second squared (newtons).

For some large nn, we might approximate

W≈∑i=1nF(xi)ΔxW \approx \sum_{i=1}^n F(x_i)\Delta x

where Δx=b−an\Delta x = \frac{b-a}{n} and xix_i is some number in the interval [a+(i−1)Δx,a+iΔx][a+(i-1)\Delta x, a+i\Delta x]. (This is just the general form of a Riemann sum).

  1. What are the units of Δx\Delta x?

  2. What are the units of F(xi)F(x_i)?

  3. Using your answers above, what are the units of WW?

Remark: we already know the units of WW 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=31+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 50 dollars10 pounds=5010 dollarspound=5 dollarspound\dfrac{50\text{ dollars} }{10\text{ pounds}} = \dfrac{50}{10}~\frac{\text{dollars}}{\text{pound}} = 5 ~\frac{\text{dollars}}{\text{pound}}. (Not 5 pound-dollars, or 5 pounds.)

Answer

(a) metres (b) newtons (c) joules

Full solution
Q4Stage 1

Suppose f(x)f(x) has units smootmegaFonzie\dfrac{\mathrm{smoot}}{\mathrm{megaFonzie}}, and xx 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\int_0^1 f(x)\,\dee{x}?

Hint

See Question 3.

Answer

smoot⋅barnmegaFonzie\dfrac{\mathrm{smoot}\cdot\mathrm{barn}}{\mathrm{megaFonzie}} (smoot-barns per megaFonzie)

Full solution
Q5Stage 1

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?

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

You may assume that the spring obeys Hooke's law.

Hint

Hooke's law says that the force required to stretch a spring xx units past its natural length is proportional to xx; that is, there is some constant kk associated with the individual spring such that the force required to stretch it xx m past its natural length is kxkx.

Answer

10 cm below the bottom of the unloaded spring

Full solution
Q6Stage 1

The work done by a force in moving an object from position x=1x = 1 to x=bx = b is W(b)=−b3+6b2−9b+4W(b) = -b^3+6b^2-9b+4 for any bb in [1,3][1,3]. At what position xx in [1,3][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=1x=1 to x=bx=b is W(b)=∫1bF(x) dxW(b) = \int_1^b F(x)~\dee{x}, where F(x)F(x) is the force on the object at position xx. To recover the equation for F(x)F(x), use the Fundamental Theorem of Calculus.

Answer

x=2x=2

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q7Stage 2Past exam · 2014D

A variable force F(x)=axF(x) = \frac{a}{\sqrt{x}} Newtons moves an object along a straight line when it is a distance of xx meters from the origin. If the work done in moving the object from x=1x = 1 meters to x=16x = 16 meters is 1818 joules, what is the value of aa? Don't worry about the units of aa.

Hint

Review Definition 2.1.1 in the CLP-2 text for calculating the work done by a force over a distance.

Answer

a=3a=3

Full solution
Q8Stage 2

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\ell, when the plunger has been pushed xx metres past its natural position, the force exerted by the air is cℓ−x\frac{c}{\ell-x} N, where cc is a positive constant (depending on the particulars of the tube of air) and x<ℓx<\ell.

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

  1. What are the units of cc?

  2. 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\ell>1.5.)

Hint

For (a), cℓ−x\frac{c}{\ell-x} is meausured in Newtons, while ℓ\ell and xx 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−1ℓ−1.5) J\displaystyle c\log\left(\frac{\ell-1}{\ell-1.5}\right)\text{ J}

Full solution

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 −9.8-9.8 metres per second squared.

Q9Stage 2Past exam · 2016Q3

Find the work (in joules) required to stretch a spring 1010 cm beyond equilibrium, if its spring constant is k=50 N/mk=50\ \mathrm{N}/\mathrm{m}.

Hint

See Example 2.1.2 in the

CLP-2 text. Be careful about your units.

Answer

14 J\dfrac{1}{4}\ \mathrm{J}

Full solution
Q10Stage 2Past exam · 2013A

A force of 1010 N (newtons) is required to hold a spring stretched 55 cm beyond its natural length. How much work, in joules (J), is done in stretching the spring from its natural length to 5050 cm beyond its natural length?

Hint

Be careful about the units.

Answer

2525 J

Full solution
Q11Stage 2Past exam · 2016Q3

A 55-metre-long cable of mass 88 kg is used to lift a bucket off the ground. How much work is needed to raise the entire cable to height 55 m? Ignore the mass of the bucket and its contents.

Hint

Suppose that the bucket is a distance yy above the ground. How much work is required to raise it an additional height dy\dee{y}?

Answer

196 J196\ \mathrm{J}

Full solution
Q12Stage 2

A tank 1 metre high has pentagonal cross sections of area 3 m2^2 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 cm3^3.

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

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
Q13Stage 2Past exam · 2015A

A sculpture, shaped like a pyramid 33m 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 zz m above ground level is a square whose side length is (3−z)(3-z) m. All of the iron plates started on the floor of a basement 22 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 9.89.8 m/{}/{}s2{}^2 for the acceleration due to gravity and 80008000 kg/{}/{}m3{}^3 for the density of iron.)

Hint

Consider the work done to lift a horizontal plate from 2 m below the ground to a height zz. You'll need to know the mass of the plate, which you can calculate from its volume, since its density is given to you.

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

Answer

∫03(9.8)(8000)(2+z)(3−z)2 dz\displaystyle\int_0^3 (9.8)(8000) (2+z)(3-z)^2\,\dee{z} joules

Full solution
Q14Stage 2

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 kk from the information about the hanging kilogram.

Answer

0.2352 J

Full solution
Q15Stage 2

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 400400 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

2049\dfrac{20}{49} kg, or about 408 grams

Full solution
Q16Stage 2

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
Q17Stage 2

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\mu \times m \times g, where μ\mu is a constant, mm is the mass of the box and gg is the acceleration due to gravity. You may assume g=9.8g=9.8 m/sec2^2.

  1. How much work is done dragging a box of mass 10 kg along the floor for three metres if μ=0.4\mu=0.4?

  2. 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 tt seconds (0≤t≤30 \leq t \leq 3) is (10−t)(10-\sqrt{t}) kilograms. If μ=0.4\mu=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 J3.92\left[30-2\sqrt{3}\right]\approx 104\text{ J}

Full solution

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.

Q18Stage 2

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?

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

You may assume that the ball starts moving with initial velocity v0v_0, 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 12mv2\frac{1}{2}mv^2, where mm is the mass of the object and vv is its velocity. Hooke's law will tell you how much work was done stretching the spring.

Answer

125\dfrac{1}{2\sqrt{5}} m/sec, or about 22.36 cm/sec

Full solution
Q19Stage 2

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/sec2^2 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

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q20Stage 3

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?

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

You may assume that 1 m3^3 of water has mass 1000 kilograms, the acceleration due to gravity is −9.8-9.8 m/sec2^2, 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\approx 0.144\text{ J}

Full solution
Q21Stage 3Past exam · 2014A

A spherical tank of radius 33 metres is half–full of water. It has a spout of length 11 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 10001000 kilograms per cubic metre. The acceleration due to gravity is 9.89.8 metres per second squared.

Figure from prob_s2.1, line 808

Figure from prob_s2.1, line 808

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

904,050π J904{,}050\pi\ \text{J}

Full solution
Q22Stage 3

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)\rho(x) = (10-x) kg/m, where xx 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\dee{x}, is (density)×\times(length) = (10−x)dx(10-x)\dee{x}, where xx 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

1020561020\frac{5}{6} J

Full solution
Q23Stage 3

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 m3^3 of water.

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 12

Figure from prob_s2.1, line 12

The force of the water acting on any tiny piece of the plunger is PAPA, where PP is the pressure of the water, and dA\dee{A} is the area of the tiny piece. The pressure varies with the depth of the piece (below the surface of the water). Specifically, P=cDP=cD, where DD is the depth of the tiny piece and cc is a constant, in this case c=9800c=9800 N/m3^3.

  1. 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.)

  2. If the length of the base is xx m, give the force of the water on the entire plunger.

  3. 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\dee{y} at depth yy.

Checking units can be a good way to make sure your calculation makes sense.

Answer

(a) 4900 N (b) 44100x2 N\frac{44100}{x^2}\text{ N} (c) 29 400 J29\,400\text{ J}

Full solution
Q24Stage 3

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 9.89.8 m/sec2^2 for the acceleration due to gravity.

Hint

When yy metres of rope have been hauled up, what is the mass of the water?

Answer

220.5 J

Full solution
Q25Stage 3

The force of gravity between two objects, one of mass m1m_1 and another of mass m2m_2, is F=Gm1m2r2F = G\dfrac{m_1m_2}{r^2}, where rr is the distance between them and GG 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 6×10246 \times 10^{24} kg and the mass of the moon is 7×10227 \times 10^{22} kg, and that they are currently 400 000400\,000 km away from each other. Also, assume G=6.7×10−11 m3kg⋅sec2G = 6.7\times 10^{-11} \ \frac{\text{m}^3}{\text{kg}\cdot\text{sec}^2}, 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 J7 \times 10^{28} \text{ J}

Full solution
Q26Stage 3

True or false: the work done pulling up a dangling cable of length ℓ\ell and mass mm (with uniform density) is the same as the work done lifting up a ball of mass mm a height of ℓ/2\ell/2.

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 9

Figure from prob_s2.1, line 9

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
Q27Stage 3

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 hh 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=x2y=x^2 and y=2−h−3x2y=2-h-3x^2. At height hh metres above the bottom of the tank, the density of the liquid is 10002−h1000\sqrt{2-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 9.89.8 m/sec2^2.

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)M(h) for the mass of a thin layer of liquid hh metres below the top of the tank, using mass==volume×\timesdensity. The rest of the problem is similar to other tank-pumping problems in this section.

Answer

925559 J9255\tfrac{5}{9}\text{ J}

Full solution
Q28Stage 3

An hourglass is 0.2 m tall and shaped such that that yy metres above or below its vertical centre it has a radius of y2+0.01y^2+0.01 m.

It is exactly half-full of sand, which has mass M=17M=\frac{1}{7} kilograms.

How much work is done on the sand by quickly flipping the hourglass over?

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 2

Figure from prob_s2.1, line 15

Figure from prob_s2.1, line 15

Assume that the work done is only moving against gravity, with g=9.8g=9.8 m/sec2^2, 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

740=0.175 J\displaystyle\frac{7}{40}= 0.175 \text{ J}

Full solution
Q29Stage 3

Suppose at position xx a particle experiences a force of F(x)=1−x4F(x) = \sqrt{1-x^4} N. Approximate the work done moving the particle from x=0x=0 to x=1/2x=1/2, accurate to within 0.010.01 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: 14[1−(18)4+1−(38)4]\displaystyle \frac{1}{4}\left[\sqrt{1-\left(\frac18\right)^4} + \sqrt{1-\left(\frac38\right)^4}\right]

Full solution

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.