A moving 3.60 kg block collides with a horizontal spring whose spring constant i

Question

A moving 3.60 kg block collides with a horizontal spring whose spring constant is 442 N/m.

A) The block compresses the spring a maximum distance of 9.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.470. What is the work done by the spring in bringing the block to rest?

B) How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?

C) What is the speed of the block when it hits the spring?

Explanation / Answer

Mass of block M = 3.6 kg
Spring constant k = 442 N/m
Compression x = 9.5 cm = 0.095 m
Coefficient of kinetic friction u = 0.47

(a)
Work done by the spring = - change in potential energy of the spring
W = - (1/2)*kx^2 = - (1/2)* 442 * 0.095^2
W = - 1.99 J

(b)
Force of friction Ff = u*Mg
Ff = 0.47 * 3.6 * 9.8
Ff = 16.58 N
Displacement = x = 0.095 m
The displacement is opposite to the force of friction
Therefore work done by friction = -Ff * x
Wf = -16.58 * 0.095
Wf = - 1.575 J

(c)
Let the speed of the block = v
Change in kinetic energy of the block from when it hits the spring to when it comes to rest
dK = 0 - 1/2 * M v^2 = -1/2 * M v^2
Total work done on the block = work done by spring + work done by friction
= -1.99 - 1.575 = -3.565
By work energy theorem,
-1/2 * M v^2 = -3.565
-1/2 * 3.6 * v^2 = -3.565
v = 1.41 m/s

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