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

work done= 2.12 kJ

Step-by-step explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

`torque =(power)/(2\pi N)`

`power=2\pi N* T`

P=
`2* \pi *2500 * 4.5`

P=70,685W

P=70.685 KW

power=
`(work done)/(time)`

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ