A 4.5kg box slides down a 4.7-m -high frictionless hill, starting from rest, acr

Question

A 4.5kg box slides down a

4.7-m -high frictionless hill, starting from rest, across a 1.9-m -wide horizontal surface, then hits a horizontal spring with spring constant 550N/m . The other end of the spring is anchored against a wall. The ground under the spring is frictionless, but the 1.9-m-long horizontal surface is rough. The coefficient of kinetic friction of the box on this surface is 0.22.

What is the speed 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

Explanation / Answer

Part A)

From conservation of energy...

KE = PE

.5mv2 = mgh (mass cancels)

(.5)v2 = (9.8)(4.7)

v = 9.60 m/s

Part B)

Some of that KE is lost due to work done by friction

W = Fd = umgd

W = (.22)(4.5)(9.8)(1.9)

W = 18.43 J

Then W = change in KE

18.43 = (.5)(4.5)(9.62 - v2)

v = 9.16 m/s

Part C)

By conservation of energy for the spring...

.5mv2 = .5kx2

(4.5)(9.16)2 = (550)(x2)

x = .829 m (82.9 cm)

Part D)

Every trip across the rough patch will lose 18.43 Joules

It has initial enegry of mgh = (4.5)(9.8)(4.7) = 207.27 J, so...

202.27/18.43 = 11.2

Thus 11 complete trips across the rough patch

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