A wooden block with mass 1.75 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.75 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 7.90 m up the incline from A, the block is moving up the incline at a speed of 7.05 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is k = 0.55. 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/s2 .

Explanation / Answer

Here ,

let the potential energy stored in the spring is U

USing conservation of energy

U = 0.5 * mv^2 + mg * d * sin(theta) + us * mg * cos(theta) * d

U = 1.75 *(0.5 * 7.05^2 + 9.8 * 7.9 * sin(35) + 0.55 * 9.8 * cos(35) * 7.90)

U = 182.2 J

the initial potential energy stored in the spring is 182.2 J

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