A 4.5 kg box slides down a 4.3-m-high frictionless hill, starting from rest, across a 2.0-m-wide horizontal surface, then hits a horizontal spring with spring constant 510 N/m. The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 2.0-m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.22. What is the speed of the box just before reaching the rough surface? What is the speed of the box just before hitting the spring? How far is the spring compressed? Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

Question
Asked Dec 14, 2019
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A 4.5 kg box slides down a 4.3-m-high frictionless hill, starting from rest, across a 2.0-m-wide horizontal surface, then hits a horizontal spring with spring constant 510 N/m. The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 2.0-m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.22. What is the speed of the box just before reaching the rough surface? What is the speed of the box just before hitting the spring? How far is the spring compressed? Including the first crossing, how many complete trips will the box make across the rough surface before coming to rest?

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Expert Answer

Step 1

According to energy conservation law, the potential energy at the top of hill is equal to the kinetic energy of box just before reaching the rough surface.

Answer: 9.18 m/s

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mgh =-nv v =/2gh = J2(9.8m/s² )(4.3m) =9.18m/s

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Step 2

The kinetic energy of box on hitting the spring is equal to the difference between the kinetic energy of box on the rough surface and the work done by friction on the rough surface.

Answer:Speed of box at the moment of hitting the spring is 8.69 m/s.

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KE - F t spring = KE soface inction factio (9.18m/s)* - (0.22)(9.8 m/s³ )( 2.0m) 1; =75.6m/s = 8.69 m/s

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Step 3

Kinetic energy of box at the moment of hitting the spring is equal to the potentia...

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KEgig = PEgring "ht-spring (4.5kg)(8.69 m/s) 510N/m = 0.8 1m

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