Question

A man pushes a 4.0 kg block against a horizontal spring, compressing the spring by 20 cm. Then the man releases the block, and the spring sends it sliding across a tabletop. It stops 90 cm from where you released it. The spring constant is 325 N/m. What is the block–table coefficient of kinetic friction?

	A.	0.47
	B.	0.97
	C.	0.57
	D.	0.37

Answer 1

Using energy conservation:

KEi + PEi + Wf = KEf + PEf

KEi = 0, since initial speed of block is zero

KEf = 0, since finally block stops after traveling 90 cm

PEi = Spring potential energy during compression = (1/2)*k*x^2

PEf = 0, since finally there is no compression in spring

Wf = Work-done by friction force = -Ff*d

Ff = Force of friction = *N = *m*g

So,

0 + (1/2)*k*x^2 - *m*g*d = 0 + 0

(1/2)*k*x^2 = *m*g*d

= Coefficient of kinetic friction = k*x^2/(2*m*g*d)

= 325*0.20^2/(2*4.0*9.81*0.90) = 0.184

Here none of the given options are correct (100% Sure about this). Please ask this issue with your TA or professor before giving negative rating.