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!!
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?
3 answers
i am not getting the correct answer
your values are wrong, but i correct it
1 answer