Question

with what acceleration a lift should move downward such that the apparent weight of person of mass 'm' is observed 5/6mg?

Answer 1

Given :

➛ Mass of person = m

➛ Apparent weight = 5/6mg

➛ Lift moves downward.

To Find :

➨ Downward acc. of lift.

SoluTion :

✭ App. weight of a man in a lift :

◕ When the lift moves upwards with acceleration a,

• R = m(g + a)

◕ When the lift moves downwards with acceleration a,

• R = m(g -a)

◕ When the lift falls freely, a = g,

• R = m(g - a) = m(g - g) = 0

◕ When the lift is at rest or moves with uniform velocity, a = 0,

• R = m(g - 0) = mg

where, R denotes apparent weight

In this Question, lift moves downwards with acceleration a.

⇒ R = m(g - a)

⇒ (5/6)mg = m(g - a)

⇒ (5/6)g = g - a

⇒ a = g - (5/6)g

⇒ a = (6g - 5g)/6

⇒ a = g/6 m/s²

Answer 2

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Explanation:

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