Question

A 15.2 kg block dragged over rough horizontal surface by constant force of 124N acting at an angle of 34.9 degrees above the horizontal. The block is displaced 37.8m. The coefficient of kinetic friction is .171. Acceleration is 9.8 m/s^2. Find the work done by the 124N force. Find the magnitude of the work done by the force of friction. Find the work done by the normal force.

          A 15.2 kg block dragged over rough horizontal surface by constant force of 124N acting at an angle of 34.9 degrees above the horizontal. The block is displaced 37.8m. The coefficient of kinetic friction is .171. Acceleration is 9.8 m/s^2. 
Find the work done by the 124N force. 
Find the magnitude of the work done by the force of friction. 
Find the work done by the normal force.
        
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Added by Harold K.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A 15.2 kg block dragged over rough horizontal surface by constant force of 124N acting at an angle of 34.9 degrees above the horizontal. The block is displaced 37.8m. The coefficient of kinetic friction is .171. Acceleration is 9.8 m/s^2. Find the work done by the 124N force. Find the magnitude of the work done by the force of friction. Find the work done by the normal force.
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Transcript

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00:01 In this question, the mass of the box is given which is 15 .2 kg.
00:11 Now the applied force is given which is 124 newton.
00:19 Then the distance moved by the box is given which is 37 .8 meter and the angle between the applied force and the horizontal surface is given which is 34 .9 degree from this we have to find the work done by the given box so here we can write this the work done by the applied force can be given by f d cos theta now the value of f is 124 d will be 37 .8 and the cost theta will be 34 .9 degree so here we will get the work done by the applied force which will be equal to 4251 .4 jule now in the second question we have to find the work done by the frictional force.
01:29 First we will find the normal force which will be equal to m g minus f sine theta by equating the equation the frictional force will be given by f s is equal to the coefficient of the friction multiplied by the normal force so this will be mu multiplied by m g minus f sine theta so we have to find the work done by the frictional force so we can write this w is equal to the in the opposite due to the frictional force work done should be in opposite direction so frictional force multiply by distance so here minus fs will be mu multiply by m g minus sine theta du multiply by d we have all of this value so here we can write this value of mu will be 0 .171 then the mass will be 15 .2 gravitational acceleration is 9 .81 minus sine theta will be 34 .9 .9 the f is and multiply by f which is 124 newton and the multiply by d which is 37 .8...
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