Question

A 10 kg box slides horizontally without friction at a speed of 1 m/s. At one point, a constant force is applied to the box in the direction of its motion. The box travels 5 m with the constant force applied. The force is then removed, leaving the box with a speed of 2 m/s. What is the magnitude of the applied force

Answer 1

Answer:

Fapp = 3 N

Explanation:

• Applying the work-energy theorem, we know that the net work done on one object by an applied force, is equal to the change in the kinetic energy of the object.
• This work is just the product of the applied force times the displacement, as follows:

       $W_{net} = F_{app} * d (1)$

• This must be equal to the change in kinetic energy:

       $\Delta K = K_{f} - K_{i} = \frac{1}2}* m* (v_{f} ^{2} -v_{i} ^{2} ) (2)$

• Equating (1) and (2),  and replacing by the givens, we can solve for Fapp, as follows:

       $F_{app} =(\frac{1}2}* m* (v_{f} ^{2} -v_{i} ^{2} ) ) / d = (5 kg* (2m/s)^{2} -1m/s^{2})/ 5 m = 3 N$

• The  magnitude of the applied force is 3 N.