## A spring balance is attached to the ceiling of a lift. When the lift is at

A spring balance is attached to the ceiling of a lift. When the lift is at rest the spring balance reads 48 N of a body hang on it. If the lift moves:
(i) downward,
(ii) upward with an acceleration of 5 m/s^2,
(iii) with a constant velocity.
What will be the reading of the balance in each case?

#### Posted by: - Mark Posting Time: - 7/17/2015 1:11:23 AM Subject: - Physics

 Answer by:- Jim D Jr. Case I: when the lift is at rest: Applying Newton’s 2nd law of motion on the hanging body, we get Spring force + gravitational force = 0 Or, -kx + mg = 0 or, kx = m*9.8 or, 49 = 9.8*m Or, m = 49/9.8 = 5kg (i) When the lift is moving downward: Applying Newton’s 2nd law of motion on the hanging body, we get mg – kx = ma or, 49 – kx = 5*5 or, kx = 49-25 = 24N (ii) When the lift is moving downward: Applying Newton’s 2nd law of motion on the hanging body, we get kx-mg = ma or, kx – 49 = 5*5 or, kx = 25+49 = 74N (iii) When lift moves with constant velocity: Here, acceleration, a = 0 Applying Newton’s 2nd law of motion on the hanging body, we get mg – kx = m*a or, 49 – kx= 0 or, kx = 49N