Answer to Question #251918 in Mechanics | Relativity for Josh

Question

Calculate the power output in watts and horsepower of a shot-- putter who takes 1.20s to accelerate the 7.27kg shot from rest to 14.0m/s, while raising it 0.800m. (5 marks)

Answer 1

According to the second Newton's law, the force that acted on the shot was:

$$F = ma$$

where $m=7.27kg$ and $a$ is the acceleration:

$$a = \dfrac{v_f}{t}$$

where $v_f = 14m/s$ and $t = 1.20s$.

The work done by the shot-putter is then:

$$W = Fd = \dfrac{mv_fd}{t}$$

where $d = 0.800m$. By definition, the power (of a constant force) is given as follows:

$$P = \dfrac{W}{t} = \dfrac{mv_fd}{t^2}\\ P = \dfrac{7.27kg\cdot 14m/s\cdot 0.8m}{(1.2s)^2} \approx 5.65\times 10^3W$$

In horsepower:

$$P \approx 7.69 hp$$

Answer. $5.65\times 10^3W$ or $7.69 hp$.