Power and Heat in Physics

Question

A 72.0 kg student walks up a 15° inclined 3.70 m ramp in 5.50 s. What will be the students power output?

Answer 1

The work in this case is the weight of the person being lifted straight up. The ramp provides a mechanical advantage but the work is the same. It's the student's weight lifted in the vertical direction.

To find the vertical distance, use sin(15°) = y/3.70 or y = 3.70sin(15°) =0.95763 m

The work is the weight (force) times that distance = (72)(9.81)(0.95763) = 676.3936 joules.

The power is the work/time = 676.3936/5.50 = 122.98 watts or (using 2 sig figs), Power = 120 watts