Answer:
work done= 2.12 kJ
Explanation:
Given
N=2500 rpm
T=4.5 N.m
Period ,t= 30 s
[tex]torque =\frac{power}{2\pi N}[/tex]
[tex]power=2\pi N\times T[/tex]
P=[tex]2\times \pi \times2500 \times 4.5[/tex]
P=70,685W
P=70.685 KW
power=[tex]\frac{work done}{time}[/tex]
work done = power * time
= 70.685*30=2120.55J
= 2.12 kJ
