ICA7 - This problem has five parts. I will state the questions to each of these five parts and any additional info in "Description" below.

Question

A 1.0 kg box is launched from rest by a compressed horizontal spring. The spring has a constant k = 500.0 N/m and is compressed by 30.0 cm. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. Kinetic friction between the box and the track is negligible on the hill, but there is a coefficient of kinetic friction of 0.60 between the box and the lower horizontal part of the track. The block remains on the track at all times. The eventual goal is to find the distance d traveled down the track.

Part a) What kinds of energy are present when the block is at the top of the hill with the spring fully compressed? Find the total energy in J.

Part b) What kinds of energy are present when the block has reached the bottom of the ramp but not yet experienced the kinetic friction force? The total energy should be unchanged from part a, so find the velocity.

Part c) How much energy does the block have after it's come to rest at the end of the track? How much work is done on the block by friction?

Part d) What is the magnitude of the friction force on the block at the bottom of the track?

Part e) Using the answers generated from parts c and d, what is the distance, d, traveled by the block?

Answer 1

It is essential here to choose the reference "0" level for the potential gravitational energy. We can conveniently assume the height h is zero when the block reaches the bottom of the hill; the height of the hill is not given, let's indicate that with H.

a) When the spring is fully compressed, the total mechanical energy H is given by the gravitational energy of the block being at height H and the elastic energy of the compressed spring: H = mgH + 1/2 k x^2 (m is the mass of the block).

b) When the block reaches the bottom of the hill, then both the elastic and gravitational energies are zero, but there is kinetic energy, since the block is moving. Because total energy is conserved:

1/2 m v^2 = mgH + 1/2 k x^2

which can in principle be solved for v (at the bottom of the ramp), assuming H is also given

c) The block has no speed anymore after it comes to rest, so it has no energy. The friction force does work Won the block which eats up all the kinetic energy 1/2mv^2 we computed in (b). So, W = -1/2mv^2

(d) The magnitude of friction force is defined as F = \mu * N, \mu = friction coefficient, N = support force of the track on the block (which in this case is equal to the block weight), so F = \mu m g = 0.60 * 1 * 9,81 N.

(e) We know the work W and we know the force. Since W = Fd for a constant force that is parallel to the displacement (d = displacement), d = W/F. (F acts along the full stretch of track that the block travels before stopping.)