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Question

# A box of bananas weighing 40.0N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40 and the coefficient of kinetic friction is 0.20.a. If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?b. What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0N to the box and the box is initially at rest?c. What minimum horizontal force must the monkey apply to start the box in motion?d. What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?e. If the monkey applies a horizontal force of 180 N, What is the magnitude of the friction force and what is the box's acceleration?

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Solution

## a. If there is no force applied to the box, the friction force exerted is 0N a/c to the newtons 3rd law of motion b.Static friction coefficient × weight of the box=6Nc. Monkey must apply horizontal force slightly more than the magnitude of the frictional force to start the motioni.e. 40N x coefficient of static friction (0.40) = 16Nd. Monkey must apply a horizontal force equal to 40N x coeff of kinetic (0.20) = 8Ne. 180N to the right, minus 8N to the left (kinetic friction x weight) = 172N to the right. as, F = ma 172=40athis implies, a = 4.3m/s2

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