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# A box of bananas weighing 40.0 N rests on a horizontal ISBN: 9780321675460 31

## Solution for problem 28E Chapter 5

University Physics | 13th Edition

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Problem 28E

A box of bananas weighing 40.0 N 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 it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N 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 18.0 N, what is the magnitude of the friction force and what is the box’s acceleration?

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Solution 28E (a).static friction is the relevant type for an object at rest,and would oppose a horizontal force.with no applied force in the horizontal there is no friction force either means 0 N. (b).The maximum possible static friction force this box surface interface could produce is fs,max= s = (0.40)(40) = 16 N. Here the applied 6.0 N is below the maximum,so static friction will negate with an opposite force of 6N. (c).The monkey would need to apply more than 16 N. fs,max= sN = (0.40)(40) = 16 N. (d).In order that move at a constant velocity,the net force on the box should be zero.the monkey has to balance the friction force by a carefully applying a force of equal magnitude and opposite in direction. f = N = (0.20)(40 N) = 8 N. | | k (e).18 N is more...

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##### ISBN: 9780321675460

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