Please show all work. The 5-lb point mass sphere system is released from test wh

Question

Please show all work.

The 5-lb point mass sphere system is released from test when the angle is 60 degrees. It swings down and strikes the 10-lb stationary box exactly in the vertical position. The coefficient of friction between the box and the floor is 0.3. The distance from the ceiling to the center of the sphere is 3 feet. Determine: The velocity of the sphere right before the impact. The velocity of the sphere and the box immediately after the collinear impact. The coefficient of restitution for the collision is 0 8. The distance that the box will travel before coming to rest.

Explanation / Answer

a) by energy conservation:

m.g.h(1 - cos 60) = 1/2.m.V^2

32.174 x 3 x (1 - cos 60) = 1/2 x V^2

V = 9.82 ft/s = velocity before impact.

b) m1.V1 + m2. V2 = m1.V1' + m2.V2'

where V1 = velocity of sphere and V2 = of box and ' shows after impact

5 x 9.82 + 0 = 5 x V1' + 10 x V2'

also V2' - V1' = e (V1 - V2)

V2' - V1' = 0.8 x (9.82 - 0 )

V1' = -1.964 ft/s = velocity of sphere before impact

and V2' = 5.892 ft/s =velocity of box

c) At stop velocity = 0

decceleration = .g = 0.3 x 32.174 = 9.6522 ft/s^2

0 = V2'2 - 2 x a xs

5.8922 = 2 x 9.6522 x s

s = 1.798 ft = distance travelled before coming to stop.

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