A wooden block with mass 1.00 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.00 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 32.0 ? (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.15 m up the incline from A, the block is moving up the incline at a speed of 6.20 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.

Calculate the amount of potential energy that was initially stored in the spring.

Take free fall acceleration to be 9.80 m/s2 .anwser in joules

Explanation / Answer

here

solve for:
1) block's GPE at B
and
2) block's KE at B
and
3) energy lost to friction {force} = Ff * 4.15 {Ff = friction force}

3rd step: set SPE = value of: 1) ans + 2) ans + 3) ans
---------------------
GPE of block at B = mgh = 1 * (9.8) * (4.15 sin 32) = 21.55 J

KE of block at B = 1/2mV^2 = (0.5) * (1) * (6.20)^2 = 19.22

Normal force of incline against block = 1 * (9.8 ) * cos 32 = 8.31

ff = friction force = (0.40) * (8.31) = 3.324

energy loss to friction = (3.324) * (4.15) = 13.8 J

SPE = 21.55 + 19.22 + 13.8 = 45.57 J

so the potential energy that is stored is 45.57 J

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