Question

A worker does 500 J of work on a 10 kg box. If the box transfers 375 J of heat to the floor through the friction between the box and the floor, what is the velocity of the box after the work has been done on it?

• A. 5 m/s
• B. 10 m/s
• C. 12.5 m/s
• D. 50 m/s
• E. 100 m/s

Answer 1

The correct option is A 5 m/s

The work-energy theorem tells us that the amount of work done on an object is equal to the amount of kinetic energy it gains, and the amount of work done by an object is equal to the amount of kinetic energy it loses. The box gains 500 J of kinetic energy from the worker’s push, and loses 375 J of kinetic energy to friction, for a net gain of 125 J. Kinetic energy is related to velocity by the formula $KE=\frac{1}{2}m{v}^2$, so we can get the answer by plugging numbers into this formula and solving for v:

$KE=\frac{1}{2}m{v}^{2}v=\sqrt{\frac{2KE}{m}}$
$v^2=\frac{2KE}{m}=\sqrt{\frac{2\left(125J\right)}{10kg}}$
$=\sqrt{25J/kg}$
$=5m/s$