Work

Question 1

A force of F(x) = 3x^2 + 2 N acts on a particle. Find the work done moving it from x=1 to x=3 meters.

Answer 1

W = [x^3 + 2x] from 1 to 3 = 33 - 3 = 30 J.

Question 2

A spring has a natural length of 20 cm. A force of 40 N is required to stretch it to 25 cm. Find the work to stretch it from 25 to 30 cm.

Answer 2

k = 40/0.05 = 800 N/m. W = integral from 0.05 to 0.10 of 800x dx = 3 J.

Question 3

A variable force F(x) = 50e^(-0.5x) Newtons is applied. Find the work done from x=0 to x=4 meters.

Answer 3

W = [-100e^(-0.5x)] from 0 to 4 = 100(1 - e^-2) J.

Question 4

A 100-ft cable weighs 4 lb/ft and hangs from a tall building. Find the work to pull the entire cable to the top.

Answer 4

W = integral from 0 to 100 of 4(100 - y) dy = 20,000 ft-lb.

Question 5

A bucket weighing 10 lb is lifted 20 ft. It contains 40 lb of water but leaks at a constant rate, losing all water just as it reaches the top. Find the work.

Answer 5

Force = 10 + (40 - 2y). W = integral from 0 to 20 of (50 - 2y) dy = 600 ft-lb.

Question 6

Find the work done by a force F(x) = 10 / (x+1)^2 from x=0 to x=9 meters.

Answer 6

W = [-10 / (x+1)] from 0 to 9 = -1 - (-10) = 9 J.

Question 7

A gas in a cylinder expands from V=1 to V=4 m^3. Pressure follows P(V) = 200/V. Find the work (W = integral P dV).

Answer 7

W = 200 ln(V) from 1 to 4 = 200 ln(4) J.

Question 8

A force F(x) = sin(pi*x / 2) Newtons is applied from x=0 to x=2. Find the work.

Answer 8

W = [-(2/pi)cos(pi*x/2)] from 0 to 2 = 4/pi J.

Question 9

A 200-lb engine is pulled 50 ft up a vertical shaft by a 5 lb/ft chain. Find the work done.

Answer 9

W = integral from 0 to 50 of (200 + 5(50 - y)) dy = 16,250 ft-lb.

Question 10

A force decreases linearly from 100 N at x=0 to 0 N at x=50. Find the work done over this distance.

Answer 10

F(x) = 100 - 2x. W = integral from 0 to 50 of (100 - 2x) dx = 2,500 J.

Question 11

A particle is moved along the x-axis by a force F(x) = x * sqrt(x^2 + 9). Find the work from x=0 to x=4.

Answer 11

Use u-sub (u = x^2+9). W = 1/3(x^2+9)^(3/2) from 0 to 4 = 125/3 - 27/3 = 98/3 J.

Question 12

A 50-kg mass is lifted 10m. Gravity is 9.8 m/s^2. A rope weighing 2 kg/m is used. Find total work.

Answer 12

W = integral from 0 to 10 of (50 + 2(10-y))*9.8 dy = 5,880 J.

Question 13

A cylindrical tank (r=2m, h=5m) is full of water (9800 N/m^3). Find the work to pump all water over the top edge.

Answer 13

W = integral from 0 to 5 of 9800 * pi * 2^2 * y dy = 490,000 * pi J.

Question 14

A force F(x) = 20 / sqrt(x) is applied from x=1 to x=4. Find the work.

Answer 14

W = [40*sqrt(x)] from 1 to 4 = 80 - 40 = 40 J.

Question 15

Find the work to move an object from x=1 to x=e against a force F(x) = ln(x) / x.

Answer 15

Use u-sub (u = ln x). W = [1/2(ln x)^2] from 1 to e = 0.5 J.

Question 16

A conical tank (radius 3ft, height 6ft at top) is full of water (62.4 lb/ft^3). Find work to pump water to the top.

Answer 16

Width r = 0.5y. W = integral from 0 to 6 of 62.4 * pi * (0.5y)^2 * (6-y) dy = 187.2 * pi ft-lb.

Question 17

A force F(x) = x * cos(x) acts on a particle. Find the work from x=0 to x=pi/2.

Answer 17

Use IBP (u=x, dv=cos x). W = [x sin x + cos x] from 0 to pi/2 = pi/2 - 1 J.

Question 18

Find the work to stretch a spring from 10cm beyond equilibrium to 20cm if k = 100 N/m.

Answer 18

W = integral from 0.1 to 0.2 of 100x dx = 1.5 J.

Question 19

A force follows F(x) = 100 / (x^2 + 1). Find the work done from x=0 to x=1.

Answer 19

W = [100 arctan(x)] from 0 to 1 = 100(pi/4) = 25 * pi J.

Question 20

A rope weighing 0.5 lb/ft is used to lift a 20 lb bucket up 40 ft. Find the work.

Answer 20

W = integral from 0 to 40 of (20 + 0.5(40 - y)) dy = 1,200 ft-lb.

Question 21

A force F(x) = 6x^2 - 4x + 5 is applied from x=1 to x=2. Find the work.

Answer 21

W = [2x^3 - 2x^2 + 5x] from 1 to 2 = (16 - 8 + 10) - (2 - 2 + 5) = 13 J.

Question 22

Find the work to pump water out of a hemispherical bowl of radius 3m if the bowl is full.

Answer 22

W = integral from 0 to 3 of 9800 * pi * (9 - y^2) * y dy = 9800 * pi * (81/4) J.

Question 23

A force F(x) = x * e^(x^2) acts on a particle from x=0 to x=1. Find the work.

Answer 23

Use u-sub (u=x^2). W = [1/2 e^(x^2)] from 0 to 1 = 0.5(e - 1) J.

Question 24

A chain of length L and total mass M is lying on the ground. Find the work to lift it so one end is at height L.

Answer 24

W = integral from 0 to L of (M/L)y dy = 1/2 ML.

Question 25

A force F(x) = 4 / x acts on a particle. Find the work from x=1 to x=e^2.

Answer 25

W = [4 ln x] from 1 to e^2 = 8 J.

Question 26

A rectangular tank (3x4m base, 2m high) is half full of water. Find work to pump it out the top.

Answer 26

W = integral from 1 to 2 of 9800 * (3*4) * y dy = 176,400 J.

Question 27

A force F(x) = 3 / (2x+1) acts from x=0 to x=4. Find the work.

Answer 27

W = [3/2 ln(2x+1)] from 0 to 4 = 1.5 ln(9) = 3 ln(3) J.

Question 28

Find the work to move an object against gravity F = GmM / r^2 from r = R to r = 2R.

Answer 28

W = [-GmM/r] from R to 2R = GmM / (2R).

Question 29

A force F(x) = x^3 - x acts from x=1 to x=2. Find the work.

Answer 29

W = [1/4 x^4 - 1/2 x^2] from 1 to 2 = (4 - 2) - (1/4 - 1/2) = 2.25 J.

Question 30

A 10m chain (5 kg/m) is hanging. Find the work to lift the bottom end to the top.

Answer 30

W = integral from 0 to 5 of 5(10 - 2y) dy * 9.8 = 1,225 J.

Question 31

What is the general formula for work?

Question 32

W = int a-b (F(x))dx, where a-b is the full distance traveled, and dx is the tiny pieces acted upon by said force.