Steven carefully places a 1.25-kg wooden block on a frictionless ramp, so that the block begins to slide down the ramp from rest. The ramp makes an angle of 55.3° up from the horizontal. Which forces below do non-zero work on the block as it slides down the ramp? How much total work has been done on the block after it slides down along the ramp a distance of 1.81 m? Nancy measures the speed of the wooden block after it has gone the 1.81 m down the ramp. Predict what speed she should measure. Now, Steven again places the wooden block back at the top of the ramp, but this time he jokingly gives the block a big push before it slides down the ramp. If the block\'s initial speed is 2.00 m/s and the block again slides down the ramp 1.81 m, what should Nancy measure for the speed of the block this time?

Step-by-step explanation:

a) The work done is equivalent to the change in potential energy of the block. The force due to gravity in the direction down the ramp does the work.

  ΔPE = mg(Δh) = (1.25 kg)(9.8 m/s²)(1.81 m·sin(55.3°) ≈ 18.23 J

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b) The kinetic energy of the block can be used to estimate the speed:

  KE = ΔPE = 1/2mv²

  v = √(2KE/m) = √(2(18.23 J)/(1.25 kg)) ≈ 5.4 m/s

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c) Steven has added kinetic energy to the block. So, the final kinetic energy it has will be ...

  initial KE + ΔPE = final KE

  final KE = (1/2)mv² + 18.23 J = (1/2)(1.25 kg)(2.0 m/s)² + 18.23 J

     = 2.5 J + 18.23 J = 20.73 J

So, the final velocity will be ...

  v = √(2(20.73)/1.25) ≈ 5.76 m/s

Comment on the question

For those of you seeking to copy the answer to "which forces do non-zero work?" we observe that there is no list of forces in this question to choose from. The best we can say is that the operative force is the component of gravity that is parallel to the ramp surface.

d. corresponding side lengths in similar triangles are proportional in length.

step-by-step explanation: