A 5.0-kg block is released from rest and allowed to slide down a 21.7 o incline

Question

A 5.0-kg block is released from rest and allowed to slide down a 21.7o incline with a kinetic friction coefficient of 0.5. After sliding 120.6 cm down the incline, it slides across a frictionless horizontal surface and encounters a spring. The far end of the spring is attached to a wall (see figure for problem 105 on page 231). The spring compresses from the initial length of 42.6 cm down to the maximum compression of 14.2 cm.

What is the spring constant of the spring?

Please check again and agian after you solve!!!

I tried for 5 times, but I could not get yet...

Thank you !

Explanation / Answer

height at incline,h = L* sin thetha = 120.6 * sin 21.7 = 44.6 m

Potential energy at top = m*g*h

energy lost against friction = frictional force * L
= miu*m*g*cos thetha * L
= 0.5*5*9.8*cos 21.7* 1.206
= 27.453 J

final spring potantial energy =initial gravitational potential energy - energy lost
0.5*K*x^2 = m*g*h -27.453
0.5*k*(0.426-0.142)^2 = 5*9.8*44.6 - 27.453
k = 53510 N/m

Answer: 53510 N/m

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