A wooden block with mass 1.50 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.50 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 31.0 (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 7.25 m up the incline from A, the block is moving up the incline at a speed of 5.45 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is k = 0.40. The mass of the spring is negligible.

Calculate the amount of potential energy that was initially stored in the spring.

Take free fall acceleration to be 9.80 m/s2

Explanation / Answer

gravitational potential energy of block at B = mgh = 1.50*9.8*7.25*sin31 = 54.89 J

kinetic energy of block at B = 1/2mV² = 1/2*1.5*5.45^2 = 22.28 J

Normal force of incline against block = 1.5*9.8*cos31 = 12.6 N

friction force =0.4*12.6 = 5.04 N

energy loss to friction = 5.04*7.25 = 36.54 J

spring potential energy = 36.54 +54.89+22.28 = 113.71 J

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