Question

A 700-kg elevator starts from rest. It moves upward for 4.50 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this time interval?

Answer 1

Answer:Average power of the elevator motor during the time interval,

Explanation:

Average power is the amount of work done per unit time.

i.e ,   -----------------eqn 1

Also, Work done,

Here, Force is provided by the motor and is acting upward direction , which causes a displacement ,S in upward direction.Therefore, = 0⁰

i.e Work done by the motor,

-------------------eqn 2

Consider the free-body diagram of the elevator,

By newtons second law,

-------------- eqn 3

So for finding acceleartion, ,

from kinematics equation,

here, initial velocity, = 0

final velocity, =1.75 m/s

time interval, = 4.50 s

eqn 3   

from kinematics equation,

here, distance travelled, has to be found out

initial velocity, = 0

acceleration ,

  time interval, = 4.50 s

From equ 1 & eqn 2,

Substituting values,