## Unit 4: Work and Energy Summary

In Unit 4, Work and Energy we learned about how forces moves another object in a direction creating work. Followed by this we learned about kinetic energy and potential energy (gravitational potential energy and elastic potential energy) this lead us to learn about mechanical energy in isolated and non-isolated system. Finally to end the unit we learned about power and how work, time and change in energy affects it. So lets get started!

**Terminology:****Energy**- Is the ability to do work**Work**- Is the measure of energy transferred when a force acting over a distance**Power**- Is the rate at which energy is transferred, used, or transformed.**Kinetic energy**- is the energy possessed by an object by virtue of its motion.**Potential energy**- is the energy possessed by an object by virtue of its position or condition**Gravitational Potential energy**- is the energy of an object due to its position in a gravitational field**Reference point**- is a point chosen from which distances are measured**Elastic potential energy**- is the potential energy stored when a spring is stretched**Mechanical energy**- the total energy in a system, total kinetic energy and potential energy stored in an object.**Isolated system**- is a frictionless system, the mechanical energy stays the same.**Non-isolated system**- has friction affecting the system, thus loses energy**Conservation of Mechanical energy**- energy is transferred from object to another (kinetic to potential) and energy is not lost and remains the same.**Efficiency**- is the ratio of energy output compared to the input of the system**Key points**- Maximum potential gravitational energy is stored at the maximum height
- The maximum amount of work is done when applied force is parallel to the displacement
- There is no work done when the force is perpendicular to the displacement
- Energy is lost due to friction, it is lost in the form of heat, sound and light.
- Unit of power is measured in Watt. One watt is equal to one joule per second

## Hooke's Law

Hooke's law of elasticity states that the amount by which a material body is stretched is linearly related to the force causing the deformation/stretch.

A chart representing Hooke's Law

A chart representing Hooke's Law

## Graph presenting Hooke's Law

We are able to calculate the constant in this graph by using the formula F = -kx and manipulating it to k = F / x. k = 6 x 10 ^ 2N / 4 x 10^ -4 cm = .015 N/m

## Kinetic energy, Potential energy and Mechanical energy

There are two main types of potential energy, gravitational potential energy and elastic potential energy. Gravitational potential energy is the energy stored in an object because of vertical height. Elastic potential energy is the energy stored in materials because of their stretching or compressing.

Most of the energy under our control is in the form of potential energy. Potential energy can be viewed as motion waiting to happen. When the motion is needed, potential energy can be changed into kinetic energy.

Here are two examples of potential energy being transformed into kinetic, as we all know potential energy is max at its greatest height, indicted with the girl holding the ball before it drops and the biker on top of the hill.

Here is an example of potential and kinetic energy changing. Roller coaster

Most of the energy under our control is in the form of potential energy. Potential energy can be viewed as motion waiting to happen. When the motion is needed, potential energy can be changed into kinetic energy.

Here are two examples of potential energy being transformed into kinetic, as we all know potential energy is max at its greatest height, indicted with the girl holding the ball before it drops and the biker on top of the hill.

Here is an example of potential and kinetic energy changing. Roller coaster

## Isolated and Non-Isolated systems

Isolated and non-isolated systems play a big role determining the total mechanical energy. In an isolated system no energy is lost due to friction, while in a non-isolated system energy is lost in the form of energy such as heat, sound and light. This is important because it helps you determine whether or not you will be adding friction in your equation.

## Work, Power and Time

This section is relatively small, it involves work, power and time to solve equation, Where P = W / T = E/T. In physics power is defined as the rate of doing work. Work hasn't have anything to do with the amount of time that the force acts to cause the displacement. We can often see work being done quickly and very slowly. Due to this power we have efficiency involved because nothing can be 100% efficient because of friction.

P = Power (W)

W = Work (J)

E = Change in energy (J)

T = Change in time (s)

P = Power (W)

W = Work (J)

E = Change in energy (J)

T = Change in time (s)

## Unit 4: Work and Energy problems

1. A force of 640N does 12 500J of work over displacement of 24.0m. What is the angle between the force and displacement.

W = ( F cos delta ) d

Cos delta = W / F x d

Cos delta = 12 500j / 24.0m x 640 N

Cos delta = .8130

Cos delta = -cos ( .8130)

Cos delta = 35.5

2. You pull a sled along a horizontal surface by applying a force of 620 N at an angle of 42.0 degrees above the horizontal. How much work is done to pull the sled 160m

W = (F cos delta) d

W = 620 N x cos (42) x 160m

W = 7.37 x 10^4 J

3. A book with a mass of 1.45kg gains 25.0 J of potential energy when it is lifted from the floor to a shelf. How high is the shelf above the floor?

Ep = mgh

h = Ep/mg

h = 25.0j / 1.45kg x 9.81 m/s^2

h = 1.76m

4. A mars rover lifts a bucket of dirt from the surface of Mars into a compartment on the rover. The mass of the dirt is .148kg and the compartment is .750m above the surface of Mars. If this action requires .400j of energy, what is the gravitational acceleration on Mars?

Ep = mgh

g = Ep/mg

g = .400J / .148kg x .750m

g = 3.6m/s ^ 2

5. A roller coaster trolley begins its journey 5.25m above the ground. As the motor tows it to the top of the first hill, it gains 4.20 x 10^5 J of gravitational potential energy. If the mass of the trolley and its passengers is 875kg, how far is the top of the hill above the ground.

Ep = mgh

h = Ep/mg

h = 4.20 x 10 ^ 5 J / 875kg x 9.81 m/s ^2

h = 48.9 m + 5.25m (original off the ground) = 52.5m

6. A winch pulls a 250 kg block up a 20.0 m long inclined plane that is tilted at an angle of 35.0 degrees to the horizontal . What chance in gravitational potential energy does the block undergo.

Ep = 250 kg x 9.81 m/s ^ 2 x 20m x sin (35)

Ep = 2.81 x 10 ^ 4 J

7. A force of 125 N causes a spring to stretch the length of .250m beyond its equilibrium position. What is the elastic potential enegy stored in the spring?

F = -kx

k = F/x

k = 125N / .250m

k = 500N/m

Ep = 1/2 kx^2

Ep = 1/2 x 500j x .250m^2

Ep = 15.65 J

8. The elastic constant for a spring is 750 N/m. How far must you stretch a spring from its equilibrium position in order to store 45.0 J

Ep = 1/2 kx^2

x = Square root of 2 x Ep / k

x = Square root of 2 x 45.0 J / 750 N/m

x = .346m

9. A 45.0kg girl pedals a 16.0 kg bike at a speed of 2.50 m/s. What is the kinetic energy of the system?

Ek = 1/2 mv^2

Ek = 1/2 61kg x 2.50 m/s ^ 2

Ek = 1.91 x 10 ^ 2 J

10. A skateboarder with a mass of 65.0 kg increases his speed from 1.75 m/s to 4.20 m/s. What is the increase in his kinetic energy?

Ek = 1/2 mv^2

Ek = 1/2 65.0kg x 1.75 m/s ^ 2

Ek = 99.53 J

Ek = 1/2 mv^2

Ek = 1/2 65.0 kg x 4.20 m/s ^ 2

Ek = 573.3 J

Ek = 573.3 J - 99.53 J = 474 J

Ek = 474 J

11. A bow that has an elastic constant of 2500 N/m is stretched to a position of .540 m from its rest position. If all of his elastic potential energy of the bow were to be transformed into kinetic energy of a 95.0g arrow, what would be the speed of the arrow?

Ep = 1/2 kx^2

Ep = 1/2 2500 N/m x .540m ^ 2

Ep = 365J

Ek = 1/2 mv^2

v = squareroot of 2 x Ek / m

v = squareroot of 2 x 365J / .095kg

v = 87.6 m/s

12. What is the speed of a 4.50 kg cannon ball if, at a height of 275 m above the ground, its mechanical energy relative to the ground is 6.27 x 10 ^ 4 J?

Ep = mgh

Ep = 4.50 kg x 9.81 m/s ^ 2 x 275m

Ep = 12139.875 J

6.27 x 10 ^ 4 J = 12139.875 J = 50560.125J

v = square root of 2 x Ek / m

v = square root of 2 x 50560.125J / 4.50 kg

v = 150 m/s

13. A force of 150 N up acts on a 9.00 kg mass lifting it to a height of 5.00 m. What is the work done on the mass by this force? What is the gravitational potential energy?

W = Fd

W = 150 N x 5.0 m

W = 750m

Ep = mgh

Ep = 5m x 9.81 m/s ^ 2 x 9.00 kg

Ep = 441J

14. In an isolated system, a crate with initial kinetic energy of 250 J and gravitational potential energy of 960J is sliding down a frictionless ramp. If the crate loses 650 J.

Ek1 + Ep1 = Ep2 + Ep2

250 J + 960 = 960 - 650

=900j

15. The engine of a crane lifts a mass of 1.50 t to a height of 65.0 m in 3.50m. What is the power output of the crane?

W = mgh

W = 1500kg x 65.0m x 9.81 m/s ^ 2

W = 956475 watts

P = W / t

P = 956475 J / 210 s

P = 4.55 KW

16. If a motor is rated 5.60 kW, how much work can it do in 20 minutes?

20minutes x 60 = 1200s

P = W / t

W = P x t

W = 5600 watts x 1200 s

W = 6.72 x 10 ^ 6 J

17. What is the power output of an electric motor that lifts an elevator with a mass of 1500 kg at a speed of .750 m/s

P = mvg

P = 1500kg x .750m/s x 9.81 m/s ^ 2

P = 11.0 kW

**Answer:**W = ( F cos delta ) d

Cos delta = W / F x d

Cos delta = 12 500j / 24.0m x 640 N

Cos delta = .8130

Cos delta = -cos ( .8130)

Cos delta = 35.5

2. You pull a sled along a horizontal surface by applying a force of 620 N at an angle of 42.0 degrees above the horizontal. How much work is done to pull the sled 160m

**Answer:**W = (F cos delta) d

W = 620 N x cos (42) x 160m

W = 7.37 x 10^4 J

3. A book with a mass of 1.45kg gains 25.0 J of potential energy when it is lifted from the floor to a shelf. How high is the shelf above the floor?

**Answer:**Ep = mgh

h = Ep/mg

h = 25.0j / 1.45kg x 9.81 m/s^2

h = 1.76m

4. A mars rover lifts a bucket of dirt from the surface of Mars into a compartment on the rover. The mass of the dirt is .148kg and the compartment is .750m above the surface of Mars. If this action requires .400j of energy, what is the gravitational acceleration on Mars?

**Answer:**Ep = mgh

g = Ep/mg

g = .400J / .148kg x .750m

g = 3.6m/s ^ 2

5. A roller coaster trolley begins its journey 5.25m above the ground. As the motor tows it to the top of the first hill, it gains 4.20 x 10^5 J of gravitational potential energy. If the mass of the trolley and its passengers is 875kg, how far is the top of the hill above the ground.

**Answer:**Ep = mgh

h = Ep/mg

h = 4.20 x 10 ^ 5 J / 875kg x 9.81 m/s ^2

h = 48.9 m + 5.25m (original off the ground) = 52.5m

6. A winch pulls a 250 kg block up a 20.0 m long inclined plane that is tilted at an angle of 35.0 degrees to the horizontal . What chance in gravitational potential energy does the block undergo.

**Answer:**

Ep = mghEp = 250 kg x 9.81 m/s ^ 2 x 20m x sin (35)

Ep = 2.81 x 10 ^ 4 J

7. A force of 125 N causes a spring to stretch the length of .250m beyond its equilibrium position. What is the elastic potential enegy stored in the spring?

**Answer:**F = -kx

k = F/x

k = 125N / .250m

k = 500N/m

Ep = 1/2 kx^2

Ep = 1/2 x 500j x .250m^2

Ep = 15.65 J

8. The elastic constant for a spring is 750 N/m. How far must you stretch a spring from its equilibrium position in order to store 45.0 J

**Answer:**Ep = 1/2 kx^2

x = Square root of 2 x Ep / k

x = Square root of 2 x 45.0 J / 750 N/m

x = .346m

9. A 45.0kg girl pedals a 16.0 kg bike at a speed of 2.50 m/s. What is the kinetic energy of the system?

**Answer:**Ek = 1/2 mv^2

Ek = 1/2 61kg x 2.50 m/s ^ 2

Ek = 1.91 x 10 ^ 2 J

10. A skateboarder with a mass of 65.0 kg increases his speed from 1.75 m/s to 4.20 m/s. What is the increase in his kinetic energy?

**Answer:**Ek = 1/2 mv^2

Ek = 1/2 65.0kg x 1.75 m/s ^ 2

Ek = 99.53 J

Ek = 1/2 mv^2

Ek = 1/2 65.0 kg x 4.20 m/s ^ 2

Ek = 573.3 J

Ek = 573.3 J - 99.53 J = 474 J

Ek = 474 J

11. A bow that has an elastic constant of 2500 N/m is stretched to a position of .540 m from its rest position. If all of his elastic potential energy of the bow were to be transformed into kinetic energy of a 95.0g arrow, what would be the speed of the arrow?

Answer:Answer:

Ep = 1/2 kx^2

Ep = 1/2 2500 N/m x .540m ^ 2

Ep = 365J

Ek = 1/2 mv^2

v = squareroot of 2 x Ek / m

v = squareroot of 2 x 365J / .095kg

v = 87.6 m/s

12. What is the speed of a 4.50 kg cannon ball if, at a height of 275 m above the ground, its mechanical energy relative to the ground is 6.27 x 10 ^ 4 J?

**Answer:**Ep = mgh

Ep = 4.50 kg x 9.81 m/s ^ 2 x 275m

Ep = 12139.875 J

6.27 x 10 ^ 4 J = 12139.875 J = 50560.125J

v = square root of 2 x Ek / m

v = square root of 2 x 50560.125J / 4.50 kg

v = 150 m/s

13. A force of 150 N up acts on a 9.00 kg mass lifting it to a height of 5.00 m. What is the work done on the mass by this force? What is the gravitational potential energy?

**Answer:**W = Fd

W = 150 N x 5.0 m

W = 750m

Ep = mgh

Ep = 5m x 9.81 m/s ^ 2 x 9.00 kg

Ep = 441J

14. In an isolated system, a crate with initial kinetic energy of 250 J and gravitational potential energy of 960J is sliding down a frictionless ramp. If the crate loses 650 J.

**Answer:**Ek1 + Ep1 = Ep2 + Ep2

250 J + 960 = 960 - 650

=900j

15. The engine of a crane lifts a mass of 1.50 t to a height of 65.0 m in 3.50m. What is the power output of the crane?

**Answer:**W = mgh

W = 1500kg x 65.0m x 9.81 m/s ^ 2

W = 956475 watts

P = W / t

P = 956475 J / 210 s

P = 4.55 KW

16. If a motor is rated 5.60 kW, how much work can it do in 20 minutes?

Answer:Answer:

20minutes x 60 = 1200s

P = W / t

W = P x t

W = 5600 watts x 1200 s

W = 6.72 x 10 ^ 6 J

17. What is the power output of an electric motor that lifts an elevator with a mass of 1500 kg at a speed of .750 m/s

**Answer:**P = mvg

P = 1500kg x .750m/s x 9.81 m/s ^ 2

P = 11.0 kW