A wooden block with mass 1.40 is placed against a compressed spring at the botto

Question

A wooden block with mass 1.40 is placed against a compressed spring at the bottom of a slope inclined at an angle of 32.0 (point ). When the spring is released, it projects the block up the incline. At point , a distance of 4.15 up the incline from , the block is moving up the incline at a speed of 5.05 and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is = 0.450. 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 .

Explanation / Answer

GPE = mgh

= 1.40(9.80)(4.15 sin 32)

= 31.39 J

KE = 1/2mV²

= (0.5)(1.40)(5.05)²

= 17.85

Normal force of incline against block

= 1.40(9.80)cos 32

= 11.44

friction force = (0.450)(11.44)

= 5.15

energy loss to friction = (5.15)(4.15)

= 21.37 J

SPE = 31.39 + 17.85 + 21.37

= 70.6 J

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