Question

A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?

Answer 1

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

$W_{\rm total}=\Delta K$

or

$W_{\rm friction}+W_{\rm spring}=0-K=-K$

where K is the block's kinetic energy at the equilibrium point,

$K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J$

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

$W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J$

Compute the work performed by friction:

$W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J$

By Newton's second law, the net vertical force on the block is

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

$W_{\rm friction}=-f(0.20\,\mathrm m)$

$\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)$

$\implies \boxed{\mu\approx0.45}$