A block is released from rest at height d = 38 cm and slides down a frictionless

Question

A block is released from rest at height d = 38 cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is 0.69. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again 0.69. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops.

(a) Give L or H, which ever is appropriate.

Explanation / Answer

This problem will be completed using the work-energy theorem.

First the block slides down a frictionless ramp where it has PE at the start. The PE = mgh and will all be converted to KE.

= m(9.8)(.38) = 3.724m J of energy

The block then slides across a plateau and friction will do negative work on it to slow it down.

W = Ffd

W = Fnd (Fn is the normal force, and on a flat surface that equals mg)

W = (.69)(m)(9.8)(.38)

W = 2.57m J

Therefore, the block lost that amount of energy

So it now has 3.724m - 2.57m J or 1.154m J of Energy

Now it slide down another ramp where it gains energy again. The amount of energy it gains will be a new mgh

That is (m)(9.8)(.19)

Added energy is 1.862m J of energy

Therefore, at the bottom of the second ramp it will have 1.862m + 1.154 m = 3.016m J of energy

The next plateau will do negative work again, just like the first one

W = Ffd

W = Fnd (Fn is the normal force, and on a flat surface that equals mg)

W = (.69)(m)(9.8)(.19)

W = 1.285m J

Therefore, the block lost that amount of energy

So it now has 3.016m - 1.285m J or 1.731m J of Energy

Finally, it can use all remaining energy to climb a new incline. When it stops it will have used all of the remaining energy, so

mgh = 1.731m (Note now m cancels, so we never needed the mass of the block)

h = 1.731/9.8

h = .176 m

It climbs to a height of .177m which is 17.7 cm

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