A wooden block with mass 1.85 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.85 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 25.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.25 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 uK= 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.

U1 = ? J

Explanation / Answer

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energy stored in spring = P.E + K.E of block + work done by friction .

P.E = mgLsin = (1.85)(9.8)(4.25)sin25 = 32.56J

K.E = 0.5(m)V2 = (0.5)(1.85)(6.70)2 = 41.52J

work done by friction = (kmgcos)L = (0.50)(1.85)(9.8)(cos25)(4.25) = 34.92J

Total energy = (34.92 + 32.56 + 41.52)J

U1 = 109J

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