Question

You push a 2.0 kg block against a horizontal spring, compressing the spring by 14 cm. then you release the block, and the spring sends it sliding across a tabletop. it stops 84 cm from where you released it. the spring constant is 290 n/m. what is the coefficient of kinetic friction between the block and the table?

Answer 1

The potential energy in the spring is given by:
$V = \frac{1}{2} kx^2$
where k is the spring constant and x is the compression of the spring.

The work done by friction is given by:
$W = \int\limits {\overrightarrow F \cdot} \,\overrightarrow{ds} = F_{friction} s = \mu Ns = \mu mgs$
where s is the sliding distance, N the normal force N = mg, m is the mass of the block, g is the gravitational acceleration and μ is the coefficient of dynamic friction.

The work done by friction must be equal to the energy provided by the spring:
$\frac{1}{2} kx^2 = \mu mgs \\ \mu = \frac{kx^2}{2mgs}$