Suppose the worker in Exercise 6.3 pushes downward at an angle of 30o below the horizontal. (a) What magnitude of force must the worker apply to move the crate at constant velocity? (b) How much work is done on the crate by this force when the crate is pushed a distance of 4.5 m? (c) How much work is done on the crate by friction during this displacement? (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? 6.3 . 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?

Solution 4E 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 k =0.25 is moved by an applied force at an angle 30 0 down the horizontal plane in such a way that, the crate moves at constant velocity. It can be pictorially represented as shown below, Step 2 of 8: To draw FBD of the crate The forces acting on the cart are, Weight downward due to gravity, W= mg Normal force upwards, N=W+ F sin N=mg+ sin Frictional force k = kN fk= (kg+ F sin) Applied force along horizontal plane, F cos=f k Given data Displacement, d= 4.5 m Coefficient of kinetic friction,k =0.25 Mass of the crate, m= 30 kg 2 Acceleration due to gravity, g=9.8m/s 0 Angle, = 30 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 plus component of applied force gets cancelled with each other. Whereas the horizontal component of applied force will be equal to the frictional force as shown in the figure above, F cos=f k Using f k Nk F cos= Nk Using N=(mg+ F sin) F cos= (kg+ F sin) Substituting =0.25, g=9.8m/s , m=30 kg, = 30 0 k 0 2 0 F cos30 =0.25(30 kg×9.8m/s + F sin30 ) 0.866F=0.25(294+0.5F) 0.866 F -0.125 F=73.5 0.741 F= 73.5 F= 99.19 N Therefore, the worker has to apply a force of 99.19 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 the angle between applied force and displacement vector is , = 30 cos =cos 30 =0.866 0 W F F d cos 30 0 Substituting F= 99.19 N ,cos 30 =0.866 and d= 4.5 m W F 99.19 N (4.5 m)(0.866) W =386 J F Therefore, the work done by the applied force is 386 J.