A wooden block with mass 1.30kg is placed against a compressed spring at the bot

Question

A wooden block with mass 1.30kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 29.0? (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 7.00m up the incline from A, the block is moving up the incline at a speed of 6.50m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is ?k = 0.50. 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 9.80m/s2 .

Explanation / Answer

work done by garvitatational force wg = (m*g*sin29*7) = 1.3*9.8*sin29*7 = -43.23 J

work done by frictional force wf = mu_k*mg*cos29*7 = 0.581.3*9.8*cos29*7 = -38.99 J...
but wg+wf = -43.23-38.99 = -82.22 J...

and wg+wf= change in Kinetic energy = (1/2)*m(v^2-u^2)..
here v = 6.5 m/s
and u is the velocity at A...

-82.22 = (1/2)*m*(6.5^2-u^2)....

u^2 = 6.5^2 + 126.5 =168.75...

u = 12.99 m/sec

initial potential energy = (1/2)*m*u^2 = 0.5*1.3*168.75 = 109.6875 J

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