Question

A 4kg block is given a small speed of 1.5 m/s down a 23(degree) incline with a coefficient of friction of 0.15. The block travels 120cm along the incline before a spring with a spring constant of 500 N/m brings the block to a momentary stop. To what maximum distance does the spring compress?

Answer 1

Concept : Work -Energy Theorem :

Steps:

Applying Newton's Second Law of motion and resolving forces

F = ma

We have :-

mgsin = X N + ma ........................(1)

N = mgcos ...........................(2)

N = Normal force acting on block = 4 x 9.8 x cos 23 = 36.08N

1) Let the compression in spring be 'x'.

Work done by Frictional Force :-

Wf = x N x 1.2m cos 180 = 0.15 x 36.08 x 1.2 = - 6.49 N

2) Work done by Spring Force :- Let the compression in spring be x m

Wspring = 1/2 x k(xi^2-xf^2)

Wspring = 1/2 x 500 x (x^2 - (1.2 - x)^2) .......................(2)

3) By Work Energy Theorem,

Total Work Done = Change in K.E

Equating Total Work Done = 1/2 m X v^2 [ v - Initial speed of block ]

and solving we get :-

2.4x = 1.48396

x = 0.6183 m [ Max compression in spring ]