Question

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 seconds.

Answer 1

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P=$2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power=$\frac{work done}{time}$

work done = power * time

                  = 70.685*30=2120.55J

                  = 2.12 kJ