(13) A 40.0kg box initially at rest is pushed 5.00m along a rough, horizontal fl

Question

(13) A 40.0kg box initially at rest is pushed 5.00m along a rough, horizontal floor with a constant applied horizontal force with a constant applied horizontal forceof 130N. The coefficient of friction between box and floor is 0.300. Find (a) the work done by the applied force,(b) the increase in internal energy in the box- floor system as a result of friction, (c) the work done by the normal force,(d) the work done by the gravitational force,(e) the change in kinetic energy of the box, and (f) the final speed of the box.

Explanation / Answer

work done by the applied force=force*displacement=130*5=650 J frictional force=40*g*0.3=117.6 N work done by frictional force=increase in internal energy in thebox-floor system as result of friction                                            =frictionalforce*displacement                                                       =117.6*5                                            =588 J normal force and the direction of displacement of the box areperpendicular to each other work done by normal force=0 weight of the box is perpendicular to the direction ofdisplacement work done by the weight of the box=0 net work done=650-588=62 J net work done = change in kinetic energy=62 J initial kinetic energy=0 after 5 m kinetic energy=62 J let the final velocity of the body=v (1/2)*40*v2=62 v=1.76 m/s final speed of the box =1.76 m/s

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