A wooden block with mass 1.65 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.65 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 33.0 ? (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.30 m up the incline from A, the block is moving up the incline at a speed of 6.85 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.

Part A Calculate the amount of potential energy that was initially stored in the spring. Take free fall acceleration to be 9.80 m/s2 . Submit My Answers Give Up

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 6.30 {ff = friction force}

3rd step: set SPE = value of: 1) ans + 2) ans + 3) ans

GPE of block at B = mgh = 1.65(9.80)(6.3 sin 33) = 55.5 J
KE of block at B = 1/2mV² = (0.5)(1.65)(6.85)² = 38.7 J
Normal force of incline against block = 1.65(9.80)cos 33 = 13.56

ff = friction force = (0.50)(13.56) = 6.78
energy loss to friction = (6.78)(6.3) = 42.714 J

SPE = 55.5 + 38.7 + 42.714 = 136.914 J

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