Question

A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressing the spring 0.123 m. After the block is released, the block moves 0.281 m to the right before coming to rest. The acceleration of gravity is 9.81 m/s 2 . What is the coefficient of kinetic friction between the horizontal surface and the block?

Answer 1

`U_k = 0.113`

Step-by-step explanation:

using the law of the conservation of energy:

`E_i -E_f=W_f`

`(1)/(2)Kx^2=NU_kd`

where K is the spring constant, x is the spring compression, N is the normal force of the block,
`U_k` is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

`(1)/(2)(138)(0.123)^2= (32.8635)U_k(0.281m)`

solving for
`U_k`:

`U_k = 0.113`