2. A 2.0 kg block slides down a 6.0-m-high frictionless hill (starting from rest

Question

2. A 2.0 kg block slides down a 6.0-m-high frictionless hill (starting from rest), crosses a 2.0-m-long horizontal surface, and then hits a horizontal spring with spring constant 300 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 block on this surface is 0.25. ???6.0 m ?????????????????????2.0 m ??

(a) What is the speed of the block just before reaching the rough surface?

(b) What is the speed of the block just before hitting the spring?

(c) How far is the spring compressed?

(d) Including the first crossing, how many complete trips will the block make across the rough surface before coming to rest?

Explanation / Answer

part a )

there is no external force so all PE is convert into KE

mgh = 1/2 * mv^2

v = sqrt(2gh) = 10.844 m/s

part b )

there is friction force = KEi - wf = KEf

f = mu*mg = 0.25 * 2 * 9.8 = 4.9

wf = 4.9 *2 = 9.8 J

117.6 - 9.8 = 1/2 * mv^2

v = 10.38 m/s

part c )

KE = 1/2 *kx^2

1/2 * mv^2 = 1/2 *kx^2

x = sqrt(KE/k)

x = 0.599 = 0.6 m = 60 cm

part d )

starting energy of box = mgh = 2 * 9.8 * 6 = 117.6 J

every time he loose energy by rough surface = mu*mg*d = 0.25 * 2 * 9.8 * 2 = 9.8 J

number of round = 117.6/9.8 = 12 trip

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