Question #251918

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)


1
Expert's answer
2021-10-18T11:14:06-0400

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


F=maF = ma

where m=7.27kgm=7.27kg and aa is the acceleration:


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

where vf=14m/sv_f = 14m/s and t=1.20st = 1.20s.

The work done by the shot-putter is then:


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

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


P=Wt=mvfdt2P=7.27kg14m/s0.8m(1.2s)25.65×103WP = \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:


P7.69hpP \approx 7.69 hp

Answer. 5.65×103W5.65\times 10^3W or 7.69hp7.69 hp.


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