A 500 kg elevator starts from rest. It moves upward for 3.0s with a constant acceleration until it reaches its cruising speed of 3 $m s^{- 1}$ . Neglect the friction. $(g = 10 m / s^{2})$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Average power of the elevator motor during the period of acceleration is 8.25 kW.

Power of the motor when it moves at its cruising speed is 15 kW.

Power of gravity remains constant throughout the motion.

Sum of the works done by all the forces on the elevator during first 4 s of its motion is 2.25 kJ.

answer is A, D, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Displacement of elevator by the time it acquires a speed of 3 $m s^{- 1}$ is calculated as
$v = u + a t \Rightarrow 3 = 0 + a \times 3$ $\Rightarrow a = 1 m s^{- 1}$
$x = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 1 \times 3^{2} = 4.5 m$
$W_{m o t o r} = Δ K + Δ U$ $= \frac{1}{2} m v^{2} + m g x$ $= \frac{1}{2} \times 500 \times 3^{2} + 500 \times 10 \times 4.5$
$= 2250 + 22500 = 24.75 k J$
$P_{m o t o r} = \frac{W_{m o t o r}}{t}$

$= \frac{24.75}{3}$ $= 8.25 k W$
When speed is constant $(v = 3 m s^{- 1})$ , force by motor = mg.
$∴ P_{m o t o r} = m g . v = 15 k W$
Power of gravity is changing as speed is changing.
$W_{A L L} = Δ K = \frac{1}{2} m v^{2} = 2.25 k J$ [ $W_{n e t}$ = 0 between 3s and 4s]