Problem 2 Two boxes are shown in the diagram above. Suppose M = 11 kg and h = 3.

Question

Problem 2

Two boxes are shown in the diagram above. Suppose M = 11 kg and h = 3.8 m.

Calculate the initial potential energy of box M.

Calculate the speed of box M immediately before colliding with box 1.5M assuming there is negligible friction acting along the ramp.

Calculate the speed of both boxes immediately after the collision if they stick together.

Calculate how far the two boxes travel after the collision if the coefficient of kinetic friction between the boxes and the horizontal surface is 0.16 after the collision.

Describe, but don’t calculate how the previous result would change if there were friction along the ramp.

Explanation / Answer

Here ,

for mass , M = 11 kg ,

the initial potential energy of box M is

Potential energy = m*g*h

Potential energy = 11 * 9.8 * 3.8

Potential energy = 409.6 J

the potential energy of the box M is 409.6 J

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now, let the speed before colliding is u

u = sqrt(2 * g * h)

u = sqrt(2 * 9.8 * 3.8)

u = 8.63 m/s

the speed of block M before colliding is 8.63 m/s

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Now , let the speed after the collision is v m/s

using conservation of momentum ,

(1.5 + M ) * v = M * 8.63

v = 3.45 m/s

the speed of box after the collision is 3.45 m/s if they stick together.

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Now , acceleration due to friction ,

a = -u*g

a = -0.16 * 9.8

a = -1.57 m/s^2

Using third equation of motion

- 3.45^2 = -2 * 1.57 * d

d = 3.8 m

the boxes slide 3.8 m

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