Spring

A 37kg object is placed onto a coil spring which has been compressed 43mm.The force used to compress the spring was 35KN.The spring and the object are on the same incline plane angled at 21deg from the horizontal. If the coefficient of friction is 0.37,how far will the block travel up the slope?
x= 0.043 m
F= 35000 N
k=F/x= 813953.49 N/m

Energy stored in spring=1/2 K x 2=
752.5000015 Joules
This energy will be dissipated in moving up the incline(gain in Potential Energy) and in overcoming friction (Mu N) and the component of weight acting down the slope (mg Sin 21)
Mu = 0.37
m= 37 Kg
Sin 21= 0.35836795
Cos 21= 0.933580426

Normal Force=N=mg Cos 21= 338.8616874
Frictional Force=Mu N 125.3788243
Component of weight acting down the slope=mg Sin 21= 130.0768146
255.455639

Work done against frictional force and component of weight
acting down the slope= 255.455639 l
Gain in PE=mg * l sin 21= 130.0768146 l
385.5324536 l

This is equal to energy stored initially in the spring= 752.5000015

Equating the two
385.5324536 l= 752.5000015
or l= 1.95 m
Distance traveled up the slope= 1.95 m

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