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# A factory worker pushes a 30.0-kg crate a distance of 4.5 ISBN: 9780321675460 31

## Solution for problem 3E Chapter 6

University Physics | 13th Edition

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Problem 3E

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Step-by-Step Solution:

Solution 3E Step 1 of 8: In the given problem, a crate of mass m=30kg is kept of the horizontal floor with coefficient of kinetic friction =0.25 is moved by an applied force in such a way that, k the crate moves at constant velocity. It can be pictorially represented as shown below, Step 2 of 8: The forces acting on the crate are, Weight downward due to gravity, W= mg Normal force upwards, N=W=mg Frictional force f = N k k Applied force, F=f k Given data Displacement, d= 4.5 m Coefficient of kinetic friction, =0.25 k Mass of the crate, m= 30 kg Acceleration due to gravity, g=9.8m/s 2 To find, The force the worker should apply, F= The work done by such force, W = F Work done by frictional force, W =f Work done by gravity, W = g Work done by normal force, W = N Total net work done on the crate, W = net Step 3 of 8: (a) What magnitude of force must the worker apply To calculate the force the worker has to apply, As the crate is moving at constant speed,the normal force and the weight gets cancelled with each other. Whereas the horizontal applied force will be equal to the frictional force as shown in the figure above, F=f k Using f = N and N=mg F= mg k k k Substituting =0.25, g=9.8m/s and m=30 kg k F=0.25(30 kg)(9.8m/s ) 2 F=f k73.5 N Since it has to overcome the friction force, worker as to apply slightly greater than frictional force. That is, F= 74 N Therefore, the worker has to apply a force of 74 N. Step 4 of 8: (b) How much work is done on the crate by this force The calculate the work done by applied force, Work done is the dot product of force and displacement, W = F.d F W = Fd cos F Where angle between force and displacement vector. Here both applied force and displacement vector are in same direction, So = 0 and cos =cos 0=1 W = F d F Substituting F=74 N and d= 4.5 m W =F74N(4.5 m) W =F33 J Therefore, the work done by the applied force is 333 J.

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Step 6 of 8

##### ISBN: 9780321675460

Since the solution to 3E from 6 chapter was answered, more than 1738 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?” is broken down into a number of easy to follow steps, and 94 words. This full solution covers the following key subjects: crate, work, done, much, Force. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 3E from chapter: 6 was answered by , our top Physics solution expert on 05/06/17, 06:07PM.

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