In: Physics

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?

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 __________.

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.

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?

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?

Part F

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

Part G

Using conservation of energy, find the speed v_{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.

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?

Part I

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

Part J

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

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.

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

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

-2mgh/3

Part D

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

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

ork 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

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

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