Question

An elevator start from rest with a constant downward acceration and occur 2.5 m in 1 sec if the weight of lift is 20 kg what should be the reaction in the rope of the lift.

Answer 1

Solution:

Given:

⇒ Initial velocity (u) = 0

⇒ Distance (s) = 2.5 m

⇒ Time (t) = 1 sec

⇒ g = 9.8 m/s²

To Find:

⇒ What should be the reaction in the rope of the lift.

Formula used:

⇒ s = ut + ½ at²

⇒ R = m(g - a)

Now, first will we calculate the value of a from 2nd equation of motion.

⇒ s = ut + ½ at²

⇒ 2.5 = 0 + ½ a × 1

⇒ 5 = a

⇒ a = 5 m/s²

Now, to find reaction in the put the value of a in 2nd formula,

⇒ R = m(g - a)

⇒ R = 20(9.8 - 5)

⇒ R = 20(4.8)

⇒ R = 96 N

Answer 2

Answer:

Explanation:

Given :-

⇒ Initial velocity of elevator = 0  (Because starts from rest)

⇒ Distance covered by elevator = 2.5 m

⇒ Time taken by elevator = 1 sec

Solution :-

From the relation s = ut + ½ at² , we get

⇒ s = ut + ½ at²

⇒ 2.5 = 0 + ½ a × 1

⇒ a = 5 m/s².

From the relation R = m(g - a). we get

⇒ R = m(g - a)

⇒ R = 20(9.8 - 5)

⇒ R = 20(4.8)

⇒ R = 96 N.

Hence, the reaction in the rope of the lift is 96 N.