**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 = 1/2mv_{i}^2 + mgh_{i} = 1/2mv_{f}^2 + mgh_{f}

Part F = K increases; U decreases; E stays the same

Part G = sqrt(v^2 + 2gh)

Part H = 1/2mv_{i}^2 + W_{nc} = 1/2mv_{f}^2

Part I = K decreases; U stays the same; E decreases

Part J = friction

Part K = 0.5mv^2 + mgh

**Solution Below:**

Part A

smooth

Part B

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

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

from the table to the floor?

Just subtract:

mgh/3 + x = -2mgh/3

x = -3mgh/3

x = -mgh

-mgh

Part E

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

1/2mv_{i}^2 + mgh_{i} = 1/2mv_{f}^2 + mgh_{f}

Part F

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

_{b}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

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

1/2mv_{i}^2 + W_{nc} = 1/2mv_{f}^2

Part I

K decreases; U stays the same; E decreases

Part J

friction

Part K

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

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.

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.