Forces/ Work 10. A 10 kg crate of delectable cupcakes (with rainbow sprinkles, o

Question

Forces/ Work 10. A 10 kg crate of delectable cupcakes (with rainbow sprinkles, of course) is being held at rest at the top of a frictionless INCLINED PLANE 8.0 m long, making an angle of 30° with the horizontal. The crate is then released from this position, slides down the frictionless inclined plane, and encounters a HORIZONTAL floor. The coefficient of kinetic friction between the crate and the floor is 0.400. The crate of cupcakes comes to rest at a distance d from the bottom of the inclined plane, as shown. 8 rn 30°

Explanation / Answer

Given,

m = 10 kg ; L = 8 m ; theta = 30 deg ; uk = 0.4 m

c)The work done by the gravity is:

Wg = m g h

h = L sin(theta)

Wg = 10 x 9.8 x 8 x sin30 = 392 J

Hence, Wg = 392 J

Work done by the normal force is:

Wn = 0 J

Work done by the frictional force on the incline is

Wf = 0 (since it is frictionless)

from conservation of energy:

1/2 m v^2 = m g h

v = sqrt (2 g h) = sqrt (2 g L sin(theta))

v = sqrt (2 x 9.8 x 8 x sin30) = 8.85 m/s

Hence, the speed at the bottom is v = 8.85 m/s

[Pls put part b for so that it can be solved and compared]

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