A wooden block with mass 2.00kg is placed against a compressed spring at the bot

Question

A wooden block with mass 2.00kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0? (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 5.45m up the incline from A, the block is moving up the incline at a speed of 5.00m/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.

Part A

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

Take free fall acceleration to be 9.80m/s2 .

Explanation / Answer

friction force=2*sin 30*9.8=9.8

for this we will conserve the energy

KE+PE(spring +box) at A=KE+PE(spring +box) at B +work don by friction

taking the hz at a to be datum

0+PE(spring) +0= .5*2*5^2 + 0 + 2*9.8*5.45*sin 30 + 2*sin 30*9.8*5.45

PE(spring)=25+53.41+53.41

=131.82 J

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