A box weighing 2000 N is to be slowly slid a distance of 20 m along a straight track. The coefficient of friction between the box and the track is 0.2 . Determine the work done by a person who pulls the box using a chain oriented at an angle with the horizontal.

Step-by-step solution

Let T be the tension in the chain and N be the normal reaction. Since the box is slid slowly, the net force on it is zero. Balancing the forces in the vertical direction: N + T = mg N = mg - T Balancing the forces in the horizontal direction: T = N Substituting the value of N : T = (mg - T ) T + T = mg T (1 + ) = mg T = mg 1 + The work done by the person is the work done by the tension force over the distance d : W = (T ) d Substituting mg = 2000 N , d = 20 m , and = 0.2 : W = 0.2 2000 20 1 + 0.2 W = 8000 1 + 5 W =