A wooden block with mass 1.35 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.35 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 27.0 degree (point A). When the spring is released, it projects the block up the incline. At point B. a distance of 5.45 m up the incline from A. the block is moving up the incline at a speed of 6.70 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is mew_k = 0.50. 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/s^2.

Explanation / Answer

Usual 1st step in solving physics probs:
SKETCH the situation showing incline & angle with block's distance, mass and weight and show all forces that act perpendicular and parallel to incline

2nd step: solve for:
1) block's GPE at B
and
2) block's KE at B
and
3) energy lost to friction {force} = ff x 5.45 {ff = friction force}

3rd step: set SPE = value of: 1) ans + 2) ans + 3) ans
---------------------
GPE of block at B = mgh = 1.35(9.80)(5.45 sin 27) = 32.73 J
KE of block at B = 1/2mV² = (0.5)(1.35)(6.70)² = 30.300
Normal force of incline against block = 1.35(9.80)cos27 = 11.78
ff = friction force = (0.50)(11.78) =5.89
energy loss to friction = (5.89)(5.45) = 32.100 J

SPE =32.73 + 30.300 + 32.100 = 95.13 J ANS

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