Question

1. a 5.0 kg object is pushed across a smooth, horizontal surface by a constant force of 15 n.

if the force does 120 j of work, what is the total increase in the objects kinetic energy?
a. 5 j
b. 75 j
c. 120 j
d. 600 j

Explanation:

Step1: Recall Work-Energy Theorem

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy, $\Delta KE = W_{net}$.

Step2: Analyze the surface condition

The surface is smooth, so there is no friction. The only work done on the object is the 120 J from the applied force, so $W_{net}=120\ \text{J}$.

Step3: Relate work to kinetic energy

By the work-energy theorem, the increase in kinetic energy equals the net work done.

Answer:

c. 120 J