Need help on part D. JUST PART D!!!! Spring and Slope with Friction A wooden blo

Question

Need help on part D. JUST PART D!!!!

Spring and Slope with Friction

A wooden block with a mass of 1.90 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 29.0 with the horizontal (refer to this as location A). When the spring is released, it projects the block up the incline. At location B, a distance of 5.25 m up the incline from A, the block is observed to be moving up the incline at a measured speed of 5.40 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is 0.55. The mass of the spring is negligible.

Part A

What is the change in gravitational potential energy for the system of the block and the Earth as the block moves from point A to point B?

47.4 J Correct

Part B

How much thermal energy will have been produced in the block and the slope as the block moves from point A to point B?

47.0 J Correct

Part C

Calculate the amount of potential energy that was initially stored in the spring. Assume that the spring potential energy is zero when the spring is relaxed.

122 J Correct

Part D

Here is one last question as a final check on your understanding of your work for this problem, looking at this problem as an example of the Conservation of Energy.

The action in this problem begins at location A, with the block resting against the uncompressed spring. The action ends at location B, with the block moving up the ramp at a measured speed of 5.40 m/s . From A to B, what has been the work done by non-conservative forces, and what has been the change in the mechanical energy of the block-Earth system (the ramp is a part of the Earth)?

Give your answer as an ordered pair, with the work by non-conservative forces first, followed by a comma, followed by the change in system mechanical energy.

WNC, E = 47.02, 50.47 it said WRONG!

Explanation / Answer

D) W(nonconservative) = workdone by friction

= fk*d*cos(180)

= -mue_k*N*d

= -mue_k*m*g*cos(25)*d

= -0.55*1.9*9.8*cos(29)*5.25

= -47.02 J

change in mechanical energy energy = workdone by non conservative force

= -47.02 J

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