Question

an elevator weight 500kg is to be lifted up at a constant velocity of 0.4m/s. what should be the maximum horse power to be motor used​

Answer 1

Answer:

2.68 HP (ANS)

Explanation:

Given -

mass of lift = 500 kg

g = 10 m/s2

velocity (v) = 0.4 m/s

displacement D in 1 sec = 0.4m

to find - HP of motor used

solution -

assuming that the motor is an ideal one-

work done = force x D

= m x g x D

= 500 x 10 x 0.4 = 2000 Nm.

time taken = 1 sec

Power = work done / time

= 2000 / 1 = 2000 W = 2kW = 2.68 HP [ 1 watt = 0.00134 HP]

hence required hp of motor = 2.68 HP (ANS)

Answer 2

Given :-

Mass of the elevator = 500 kg

Velocity of the elevator = 0.4 m/s

To Find :-

The maximum horse power to be motor used​.

Analysis :-

Here we are given with the mass and the velocity of the elevator.

In order to find the power substitute the given values in the question such that power is equal to work done divided by time.

Solution :-

We know that,

• m = Mass
• w = Work done
• g = Gravity
• v = Velocity
• t = Time

Using the formula,

$\underline{\boxed{\sf Power=\dfrac{Work \ done}{Time \ taken} }}$

Given that,

Work done = mgD

Time (t) = 1

Gravity (g) = 10 m/s

Substituting their values,

⇒ p = mgD/t

⇒ p = 500 × 10 × 0.4/1

⇒ p = 2000 W

By converting,

1 W = 0.0013 Hp

2000 W = 2000 × 0.0013

= 2.68 Hp

Therefore, the maximum horse power to be motor used is 2.68 Hp.