Question

A block of mass m = 4 kg slides along a horizontal table when it encounters the free end of a horizontal spring of spring constant k = 14 N/m. The spring is initially on its equilibrium state, defined when its free end is at x=0 . Right before the collision, the block is moving with a speed vi = 5 m/s . There is friction between the block and the surface. The coefficient of friction is given by μ = 0.86 . How far did the spring compress when the block first momentarily comes to rest?

Answer 1

KE = (1/2)(m)(v)^2
= (0.5)(4)(5)^2
= 50 J

Constant friction force encountered:

N = (0.94)(4)(10)
= 37.6 N

Energy absorbed by friction:

Fd = 37.6x

Energy absorbed by the spring:

=(1/2)(17)(x)^2
=8.5x^2

Final Equation:

50 = 8.5x^2+37.6x
0 = 8.5x^2+37.6x-50

Solve the quadratic you will get the answer!!

Answer 2

i am not getting the correct answer

Answer 3

your values are wrong, but i correct it