Example-1
An elevator can carry a maximum load of 1000 kg (elevator + passengers) is moving up with a constant speed of 2 m/s. The
frictional forces opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts ?
Solution
The downward force on the elevator is
$F = m g + f_{fric}$
$f = (1000 \times 10) + 4000 = 14000 N$
The motor must supply enough power to balance this force. Hence,
$P = F. v = 14000 \times 2 = 28000 W$
Example-2
A truck of mass 1000 kg accelerates on the level road from 10 m/s to 20 m/s in 5 sec. The frictional force present is 4000N. Find the average power delivered by the engine.?
Solution
acceleration of truck = $\frac {20 -10}{5} = 2 m/s^2$
So total force delivered by the engine = $ma + f_{fric} = 1000 \times 2 + 4000 = 6000 N$
Average power = F X average velocity = 6000 X 30 = 18000W
