A shot-putter accelerates a 7.3 kg shot from rest to 14 m/s. If this motion takes 1.5 s, what average power w

Answer 1

power is in unites joules / sec. the joules of energy goes from 0 to (1/2)(7.3)(14^2) which is kinetic energy. Thus producing [(1/2)(7.3)(14^2)] / (1.5) watts of power.

Divide that by 1.5 if you want watts / second.

Answer 2

Work = fd = force X distance the force is applied. Force = f = ma; where m = 7.3 kg and a = v/t where v = 14 m/sec and t = 1.5 sec. The force f is applied d = 1/2 at^2 = 1/2 (v/t) t^2 = 1/2 vt. Thus, pulling it all together, we have W = fd = mad = m (v/t) 1/2 vt = 1/2 mv^2, which, darn, if that isn't the formula for kinetic energy we just derived.

Since power is nothing more or less than energy expended over a given time, we have P = W/t = 1/2 mv^2/t; where m = 7.3, v = 14 m/sec, and t = 1.5 sec. You can do the math.

There are some fancy names for power, given to honor arcane scientists of the past. But I prefer using the good old kg-m-sec SI units. Thus, power is in Newton-meter/sec; where Newton is in kg-m/sec^2 units.