The 3.00-kg object in the figure is released from rest at a height of 5.00 m on
ID: 2241373 • Letter: T
Question
The 3.00-kg object in the figure is released from rest at a height of 5.00 m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant 400 N/m. The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. (a) Find x. (b) Describe the motion object (if any) after the block momentarily comes to rest? [Answer: (a) 0.858 m (b) The energy stored in the compressed spring will accelerate the block, launching it back up the incline and the block will retrace its path, rising to a height of 5.00 m.]Explanation / Answer
Assuming air resistance is negligible, all of the potential energy that the object has at the top of the ramp is converted into kinetic energy by the time it gets to the bottom of the ramp. This is because no matter what path the object takes to move the 5m vertically (ie. falling straight down v. sliding on the ramp), gravity does the same amount of work on it.
Thus, calculate the total amount of potential energy at the top of the ramp:
Ep=mgh
Ep=3(9.81)5
Ep=147.15 Joules
Because all of this potential energy is converted into kinetic energy in the object by the bottom of the ramp, the object hits the spring with 147.15 J of energy.
By using the formula for elastic potential energy, you can calculate exactly how far the spring compresses.
147.15 =(1/2)k(x^2)
294.3=(400)(x^2)
0.73575=x^2
sqrt(0.73575)=x
x=0.8577 m
b. after the object compresses the spring fully and stops momentarily, the spring converts it's elastic potential energy back into kinetic energy in the object and pushes it away again.
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