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

So coefficient of kinetic friction will be equal to 0.4081

Step-by-step 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=(1)/(2)kx^2=(1)/(2)* 100* 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* s=2`

`\mu* 0.5* 9.8* 1=2`

`\mu =0.4081`

So coefficient of kinetic friction will be equal to 0.4081