A block with mass of 0.5 kg is forced against a horizontal spring of negligible mass. compressing the spring a distance of 0.20m. When released, the block moves on a horizontal tabletop for 1 meter before coming to rest. The spring constant is 100N/m. What is the coefficient of friction between the block and the tabletop??

Required"mu" is given by:

mu0.59.81 = 0.5100*0.20^2 or
mu = 4/9.8 = 0.408 or 0.41

Therefore, the coefficient of friction between the block and the tabletop is 0.408 or rounded off to 0.41.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

Answer Link

aristocles aristocles · Answer 2 · 2019-10-21T21:30:09-04:00

Answer:

\(\mu = 0.408\)

Explanation:

Here we know that work done by all force must be equal to change in kinetic energy of the block

so here we will have

\(W_{friction} + W_{spring} = K_f - K_i\)

here we know that

\(W_{friction} = -\mu mg d\)

\(W_{spring} = \frac{1}{2}kx^2\)

also we know that initially and finally block is at rest

so we have

\(\frac{1}{2} kx^2 - \mu mg d = 0 - 0\)

\(\frac{1}{2}(100)(0.20^2) - \mu (0.5)(9.81)(1) = 0\)

\(\mu = 0.408\)