A wooden block with mass 1.50 kg is placed against a com- pressed spring at the

Question

A wooden block with mass 1.50 kg is placed against a com- pressed spring at the bottom of an incline of slope 30.0 degree (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.00 m up the incline from A, the block is moving up the incline at 7.00 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is mu_k = 0.50. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

Explanation / Answer

GPE of block at B = mgh = 1.50(9.80)(6 sin 30) = 44.1 J
KE of block at B = 1/2mV² = (0.5)(1.50)(7)² = 36.75
Normal force of incline against block = 1.50(9.80)cos 30 = 12.73
ff = friction force = (0.50)(12.73) = 6.3652
energy loss to friction = (6.3652)(6) = 38.19 J

SPE = 44.1 + 36.75 + 38.19 = 119.04 J

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