A wooden block with mass 1.25 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.25 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 28.0 (point A ). When the spring is released, it projects the block up the incline. At point B , a distance of 4.90 m up the incline from A , the block is moving up the incline at a speed of 5.35 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.

Explanation / Answer

Workdone by friction = change in mechanical energy

     = final gravitational and kinetic energy - initial elastic potential energy

N*mue_k*d*cos(180) = (m*g*h + 0.5*m*v^2) - initial elastic potential energy -

initial elastic potential energy = m*g*h + 0.5*m*v^2 + N*mue_k*d

= 1.25*9.8*4.90*sin(28) + 0.5*1.25*5.35^2 + 1.25*9.8*cos(28)*0.55*4.90

= 75.22 J <<<<<<<<-----Answer

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