Question

a spring balance attached to the ceiling of a moving elevator indicated the weight of abody which weight 980N on the ground as 1470N. find both diRECTION AND MAGNITUDE OF THE ELEVATOR?

Answer 1

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 

Hope this helps

Answer 2

sorry dont know the ans