Question

A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. When released, the block moves on a horizontal tabletop for 1.0m before coming to rest. The force constant k is 100 N/m. What is the coefficient of kinetic friction between the block and tabletop

Answer 1

Answer:

So coefficient of kinetic friction will be equal to 0.4081

Explanation:

We have given mass of the block m = 0.5 kg

The spring is compressed by length x = 0.2 m

Spring constant of the sprig k = 100 N/m

Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to $W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J$

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

So $\mu mg\times s=2$

$\mu\times 0.5\times 9.8\times 1=2$

$\mu =0.4081$

So coefficient of kinetic friction will be equal to 0.4081