Download worksheet: kinetic and potential energy problems

WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS 1. Stored energy or energy due to position is known as Potential energy. 2. The formula for calculating potential energy is mgh. 3. The three factors that determine the amount of potential energy in an object are mass, gravitational acceleration, and height of an object. 4. Potential energy is measured in units of joules(J). 5. Mass must be measured in units of kg. 6. Gravitational pull must be measured in units of m/s2. 7. Height must be measured in units of m. 8. Calculate the potential energy of a rock with a mass of 55 kg while sitting on a cliff that is 27 m high. PE = mgh
In this problem, m = 55kg and h = 27m. G or gravitational acceleration is always 9.8 m/
s2.
So, PE = 55kg ⋅ 9.8 m/s2 ⋅ 27m = 14553 J
∴ 14553 J
9. What distance is a book from the floor if the book contains 196 Joules of potential energy and has a mass of 5 kg? PE = mgh
In this problem, m = 5kg and PE = 196 J. G or gravitational acceleration is always 9.8
m/s2.
So, 196 J = 5kg ⋅ 9.8 m/s2 ⋅ xm, x = 4m
∴ 4m
10. An automobile is sitting on a hill which is 20 m higher than ground level. Find the mass of the automobile if it contains 362,600 J of potential energy. PE = mgh
In this problem, h = 20m and PE = 362,600 J. G or gravitational acceleration is always
9.8 m/s2.
So, 362,600 J = xkg ⋅ 9.8 m/s2 ⋅ 20m, x = 1850kg
∴ 1850kg
11. Energy of motion is known as Kinetic energy. 12. The formula for calculating kinetic energy is 1/2mv2. 13. The two factors that determine the amount of kinetic energy in an object are mass and velocity. 14. Kinetic energy is measured in units of joules(J). 15. Mass must be measured in units of kg. 16. Velocity must be measured in units of m/s. Energy, Work and Power 17. Calculate the kinetic energy of the rock in problem #8 if the rock rolls down the hill with a velocity of 8 m/s. KE = 1/2mv2
In this problem, m = 55kg and v = 8 m/s.
So, KE = 1/2 ⋅ 82 m/s ⋅ 55kg = 1760 J
∴ 1760 J
18. Calculate the kinetic energy of a truck that has a mass of 2900 kg and is moving at 55 m/s. KE = 1/2mv2
In this problem, m = 2900kg and v = 55 m/s.
So, KE = 1/2 ⋅ 552 m/s ⋅ 2900kg = 4386250 J
∴ 4386250 J
19. Find the mass of a car that is traveling at a velocity of 60 m/s North. The car has 5,040,000 J of kinetic energy. KE = 1/2mv2
In this problem, KE = 5,040,000 J and v = 60 m/s.
So, 5,040,000 J = 1/2 ⋅ 602 m/s ⋅ xkg, x = 2800kg
∴ 2800kg
20. How fast is a ball rolling if it contains 98 J of kinetic energy and has a mass of 4 kg? KE = 1/2mv2
In this problem, KE = 98 J and m = 4 kg.
So, 98 J = 1/2 ⋅ x2 m/s ⋅ 4kg, x = 7 m/s
∴ 7 m/s
WORKSHEET: POTENTIAL ENERGY PROBLEMS Fill in the Blank: 1. Potential energy is the energy matter has as a result of its mass or height. 2. The more mass an object has the (more / less) potential energy it has. 3. The potential energy an object has due to its position is called gravitational potential energy. 4. The formula for calculating gravitational potential energy is PE = mgh. 5. The value of the g constant (the acceleration of all objects due to gravity) on earth is
9.8 m/s2.
6. The SI (metric) unit for energy is joules and the symbol is J. Sample Problems: 7. PE = 205.8 m = 0.6 kg
g = 9.8 m/s2 h = 35 m 8. PE = 30 J
m = 0.3kg
g = 9.8 m/s2 h = 10m
9. PE = 7.5 J m = 1.6 kg g = 9.8 m/s2 h = 0.48m
10. PE = 431.2 J
m = 2 kg g = 9.8 m/s2 h = 22m
11. A 10 kg mass is lifted to a height of 2 m. What is its potential energy at this position? Given Formula Substitution Answer (with units) PE = mgh
In this problem, m = 10kg and h = 2m. G or gravitational acceleration is always 9.8 m/s2.
So, PE = 10kg ⋅ 9.8 m/s2 ⋅ 2m = 196 J
∴ 196 J
12. At what height is an object that has a mass of 16 kg, it its gravitational potential energy is 7500J? PE = mgh
In this problem, m = 16kg and PE = 7500 J. G or gravitational acceleration is always 9.8
m/s2.
So, 7500 J = 16kg ⋅ 9.8 m/s2 ⋅ xm = 47.83 m
∴ 47.83 m
13. PE = 1176 m = 3 kg
g = 9.8 m/s2 h = 40 m 14. PE = 74 J
m = 3.8 kg
g = 9.8 m/s2 h = 1.99 m
15. PE = 52 J m = 0.3 kg g = 9.8 m/s2 h = 18 m
16. PE = 1078 J
m = 5 kg g = 9.8 m/s2 h = 22m 17. What potential energy is acquired by a hammer with a mass of 0.75 kg when raised 0.35m? PE = mgh
In this problem, m = 0.75 kg and h = 0.35m. G or gravitational acceleration is always 9.8
m/s2.
So, PE = 0.75kg ⋅ 9.8 m/s2 ⋅ 0.35m = 2.57 J
∴ 2.57 J
18. A book with a mass of 1 kg is dropped from a height of 3 m. What is the potential energy of the book when it reaches the floor? PE = mgh
In this problem, m = 1 kg and h = 3 m. G or gravitational acceleration is always 9.8 m/s2.
So, PE = 1kg ⋅ 9.8 m/s2 ⋅ 3m = 29.4 J
∴ 29.4 J
19. At what height is an object that has a mass of 50 kg, if its gravitational potential energy is 9800 J? PE = mgh
In this problem, m = 50 kg and PE = 9800 J. G or gravitational acceleration is always
9.8 m/s2.
So, 9800 J = 50kg ⋅ 9.8 m/s2 ⋅ xm, x = 20m
∴ 20m
20. What is the mass of an object if its gravitational potential energy is 3822 J and it is 15m above the ground? PE = mgh
In this problem, h= 15 m and PE = 3822 J. G or gravitational acceleration is always 9.8
m/s2.
So, 3822 J = x kg ⋅ 9.8 m/s2 ⋅ 15m, x = 26 kg
∴ 26 kg
21. An object with a mass of 20 kg and potential energy of 584 J is what distance above the ground? PE = mgh
In this problem, m= 20 kg and PE = 584 J. G or gravitational acceleration is always 9.8
m/s2.
So, 584 J = 20 kg ⋅ 9.8 m/s2 ⋅ xm, x = 2.98m
∴ 2.98m
Work Write the equation and units for work: Equation: W = Fd, unit is joules(J).
1. A weight lifter lifts a set of weights a vertical distance of 2 m. If a constant net force of 350 N is exerted on the weights, what is the net work done on the weights? W = Fd
In this problem, f = 350 N and d = 2m.
So, W = 350 N ⋅ 2m = 700 J
∴ 700 J
2. A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 degrees downward from the horizontal. Find the work done by the shopper on the cart as the shopper moves along a 50 m length of aisle. 3. If 2 J of work is done in raising a 180 g apple, how far is it lifted? W = Fd
In this problem, W = 2 J. Because F = mg, F = 0.18 kg ⋅ 9.8 = 1.764 N
So, 2 J = 1.764 N ⋅ xm, x = 1.13 m
∴ 1.13 m
4. For each of the following cases, indicate whether the work done on the second object in each example will have a positive or a negative value. a. The road exerts a friction force on a speeding car skidding to a stop. "
Negative value
b. A rope exerts a force on a bucket as the bucket t is raised up a well. "
Positive value
c. Air exerts a force on a parachute as the parachutist falls to Earth. "
Negative value
5. If a neighbor pushes a lawnmower four times as far as you do but exerts only half the force, which one of you does more work and by how much? W = Fd
Let’s say a neighbor pushes a lawnmower 4m with 1 N of power. Then, I would push it
1m with 2 N of power.
A neighbor did 4m ⋅ 1N = 4J of work and I did 1m ⋅ 2N = 2J of work. Thus, a neighbor
did twice more work than me.
∴ A neighbor did twice more work than me.
6. A worker pushes a 1500 N crate with a horizontal force of 345 N a distance of 24 m. Assume the coefficient of kinetic friction between the crate and the floor is 22.
a. How much work is done by the worker on the crate? b. How much work is done by the floor on the crate? c. What is the net work done on the crate? 7. A 0.075 kg ball in a kinetic sculpture moves at a constant speed along a motorized vertical conveyor belt. The ball rises 1.32 m above the ground. A constant frictional forc
e of 35 N acts in the direction opposite the conveyor belt’s motion. What is the net work done on the ball? 8. For each of the following statements, identify whether the everyday or the scientific meaning of work is intended. a. Jack had to work against time as the deadline neared.
"
Everyday work
b. Jill had to work on her homework before she went to bed. "
Everyday work
c. Jack did work carrying the pail of water up the hill. "
Scientific work
9. Determine whether work is being done in each of the following examples: a. a train engine pulling a loaded boxcar initially at rest "
Work is being done.
b. a tug of war that is evenly matched "
Work is NOT being done.
c. a crane lifting a car "
Work is being done.
10. How much work does Bobby perform in pushing a 35 N crate a distance of 4 meters
? List known values formula substitution answer & units W = Fd
In this problem, f = 35 N and d = 4m.
So, W = 35 N ⋅ 4m = 140 J
∴ 140 J
11. How far will a 70 N crate be moved if 3500 J or work are accomplished? list known values formula substitution answer & units W = Fd
In this problem, f = 70 N and W = 3500 J.
So, 3500 J = 70 N ⋅ xm, x = 50 m
∴ 50m
12. What force is needed to move a barrel 25­m if 225 J of work are accomplished? W = Fd
In this problem, d= 25 m and W = 225 J.
So, 225 J = x N ⋅ 25m, x = 5 N
∴5N
13. Peggy uses a force of 40 N to move the grocery basket 18 meters. How much work
did she perform? W = Fd
In this problem, F = 40 N and d = 18 m.
So, W= 40 N ⋅ 18m = 720 J
∴ 720 J
14. How far will a 150 N crate be moved if 600 J or work are performed? W = Fd
In this problem, f = 150 N and W = 600 J.
So, 600 J = 150 N ⋅ xm, x = 4 m
∴ 4m
15. What distance is needed to lift a box that weighs 300 N if 15000 J of work are
accomplished? W = Fd
In this problem, F = 300 N and W = 15000 J.
So, 15000 J = 300 N ⋅ xm, x = 50 m
∴ 50 m
16. How far will a 700 N crate be moved if 2800 J or work are accomplished? W = Fd
In this problem, f = 700 N and W = 2800 J.
So, 2800 J = 700 N ⋅ xm, x = 4m
∴ 4m
17. What force is needed to move a barrel 45­m if 3600 J of work are accomplished? W = Fd
In this problem, d= 45 m and W = 3600 J.
So, 3600 J = x N ⋅ 45m, x = 80 N
∴ 80 N
18. How much work does Billy perform if he pushes the 8000 N stalled car a distance? of 25 meters? W = Fd
In this problem, F = 8000 N and d = 25 m.
So, W= 8000 N ⋅ 25m = 200000 J
∴ 200000 J
19. Grant stands 3 m from the check out at the grocery store holding a 20 N bag of potatoes for 10 minutes. How much work does he perform? W = Fd
In this problem, F = 20 N and d = 3 m.
So, W= 20 N ⋅ 3m = 60 J
∴ 60 J