A wooden block with mass 1.45 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.45 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 27.0 (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.05 m up the incline from A, the block is moving up the incline at a speed of 5.00 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 .

Explanation / Answer

P.E. = (1/2)mv^2 + mg(cos + sin)s
P.E. = (1.45 kg){(1/2)(5.00 m/s)^2 + (9.80 m/s^2)[(0.40/2)3 + 1/2](6.05m)}
That's solve
P.E. = 90.89J

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