A wooden block with mass 1.55 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.55 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 31.0 degrees . When the spring is released, it projects the block up the incline. At point , a distance of 5.25m up the incline from , the block is moving up the incline at a speed of 6.15 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is = 0.45. The mass of the spring is negligible.

Calculate the amount of Potential Energy that was initially stored in the spring.

(135 J is not the answer :) )

Explanation / Answer

By conservation of energy,
the total potential energy stored in the spring gets converted into the K.E+increase in
P.E+work done by the frictional force.

so, K.E=1/2*m*v^2=(1/2)*1.55*6.15^2=29.3J
increase in gravitational P.E =mg*5.25*sin31=41.1J
work done by frictional force,Wf=mgcos31*5.25=.45*1.55*9.8*cos31*5.25=30.8J

so total energy stored in the spring=(29.3+41.1+30.8)J=101.2J

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