Question

A 650kg elevator starts from rest .It moves upward for 3s with constant acceleration untill it reaches its its cruising speed 1.75 m/s.

(a)What is the avg. power of the elevator motor during this period?

(b)How does this power comparewith its power when it moves with its cruising speed?

Answer 1

a)The acceleration of the elevator,
a=1.753=0.583 m/s2
And the height achieved by the elevator,
h=12at2=12×0.583×32=2.625 m
Work done in this process
W=m(g+a)h=650×(10+0.583)×2.625=18057.24 J
Average power,
P=Wt=6019.08 watt
b) Power needed for the crusing speed,
P=Fv=mgv=650×10×1.75=11375 watt
 

Answer 2

The average power and the power at cruising speed of the elevator will be equal to 6019 W and 11375 W respectively.

Given:

Mass (m) = 650 Kg

Initial Velocity (u) = 0 m/s

Final Velocity (v) = 1.75 m/s

Time (t) = 3 s

To Find:

The average power and the power when the elevator move at its cruising speed.

Solution:

It's fairly simple to find the answer to this question, as seen below.

We know that, according to the first law of motion,

Acceleration = $\frac{Final Velocity - Initial Velocity}{Time}$

$a = \frac{v - u}{t}$

On substituting the values,

a = $\frac{1.75 - 0}{3}$

a = 0.583 m/s²

Then, the height achieved by the elevator,

h = ut + ½ at²

We know that u = o, hence ut = 0

So,

h = ½ at²

h = ½ (0.583) × (3²)

h = 2.625 m

We know that,

Work done (W) = m (g+a) h

On substituting the values,

W = 650 × (10+0.583) × 2.625

W = 18057.5 J

Then,

Average power = $\frac{Work}{Time}$

$P_{avg}$ = $\frac{W}{t}$

$P_{avg}$ = $\frac{18057.5}{3}$

$P_{avg}$ = 6019 W

Hence, the average power of the elevator = 6019 W

Now,

Power at cruising speed = Force x Velocity

P = Fv

P = mgv

On substituting the known values,

P = 650 × 10 × 1.75

P = 11375 W

Hence, the power at cruising speed will be 11375 W

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