## Mastering Physics Solutions: Where’s the Energy?

Where’s the Energy?

Part A = smooth
Part B = a distance 2h/3 above the floor
Part C = -2mgh/3
Part D = -mgh
Part E = 1/2mvi^2 + mghi = 1/2mvf^2 + mghf
Part F = K increases; U decreases; E stays the same
Part G = sqrt(v^2 + 2gh)
Part H = 1/2mvi^2 + Wnc = 1/2mvf^2
Part I = K decreases; U stays the same; E decreases
Part J = friction
Part K = 0.5mv^2 + mgh

Solution Below:

In this problem, we will consider the following situation as depicted in the diagram : A block of mass m slides at a speed v along a horizontal, smooth table. It next slides down a smooth ramp, descending a height h, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor). You will analyze the motion of the block at different moments using the law of conservation of energy.

Part A

Which word in the statement of this problem allows you to assume that the table is frictionless?

smooth

Part B

Suppose the potential energy of the block at the table is given by mgh/3. This implies that the chosen zero level of potential energy is __________.

Since U = mgh, we are “1/3 h” above the zero level. That means that “2/3h” must be zero. This is difficult to understand conceptually, and difficult to explain in words without using pictures and walking you through step by step- your professor should be able to help.

a distance 2h/3 above the floor

Part C

If the zero level is a distance 2h/3 above the floor, what is the potential energy of the block on the floor?
Express your answer in terms of some or all the variables m, v, h and and any appropriate constants.

Since the block is below the “zero level,” it will have a negative potential energy.

-2mgh/3

Part D

Considering that the potential energy of the block at the table is mgh/3 and that on the floor is -2mgh/3, what is the change in potential energy ΔU of the block if it is moved
from the table to the floor?

Just subtract:

mgh/3 + x = -2mgh/3
x = -3mgh/3
x = -mgh

-mgh

Part E

Which form of the law of conservation of energy describes the motion of the block when it slides from the top of the table to the bottom of the ramp?

Overall energy is conserved. The proportion of kinetic energy to potential energy changes, but the total amount of energy remains the same.

1/2mvi^2 + mghi = 1/2mvf^2 + mghf

Part F

As the block slides down the ramp, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?

Going down the ramp, potential energy is converted to kinetic, but the overall amount of energy remains the same, because energy is conserved.

K increases; U decreases; E stays the same

Part G

Using conservation of energy, find the speed vb of the block at the bottom of the ramp.
Express your answer in terms of some or all the variables m, v, h and and any appropriate constants.

PE(i) + KE(i) = PE(f) + KE(f)

At the top of the ramp, potential energy will be at a maximum and kinetic energy will be at a minimum. At the bottom of the ramp, kinetic energy will be at a maximum and potential energy will be minimum. But although the type of energy changes, the total amount of energy remains the same:

mgh + 1/2mv^2 = mgh(b) + 1/2mv(b)^2

Since the height at the bottom is zero, we can eliminate potential energy from the right side of the formula:

mgh + 1/2mv^2 = 1/2mv(b)^2
gh + 1/2v^2 = 1/2v(b)^2
2gh + v^2 = v(b)^2
v(b) = sqrt(v^2 + 2gh)

sqrt(v^2 + 2gh)

Part H

Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops?

Work is the change in energy, so the effect of friction is to perform work.

1/2mvi^2 + Wnc = 1/2mvf^2

Part I

As the block slides across the floor, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?

K decreases; U stays the same; E decreases

Part J

What force is responsible for the decrease in the mechanical energy of the block?

friction

Part K

Find the amount of energy E dissipated by friction by the time the block stops.
Express your answer in terms of some or all the variables m, v, h and and any appropriate constants.

Friction will slow the block to zero, so it will dissipate all of the kinetic and potential energy that the block has at the top of the ramp.

0.5mv^2 + mgh

### 2 Responses to Mastering Physics Solutions: Where’s the Energy?

1. Kristi Beattie says:

Someone please show me step by step how the answers are being found, I can not seem to get the tutor to show me I have the answers but wish to know how they are achieved. I do not know the steps , somehow I am missing something.

• Mastering Physics Solutions says:

We’ve added some extra explanation that may help you. This problem presents some very fundamental concepts though, so if you are still having difficulties we recommend talking with your professor for additional guidance.