Question

A Spring Balance Is Attached To The Ceiling Of A Lift. A Man Hangs His Bag On The Spring And The Spring Balance Reads 49N, When The Lift Is Stationary. If The Lift Moves Downward With An Acceleration Of 5 Ms-2, The Reading Of The Spring Balance Will Be______

Answer 1

Answer:

As we learnt in

Lift is moving down with a = g -

a=g

mg-R=mg

R=0

- wherein

Apparent weight = 0 (weightlessness)

 

 

 

When lift is standing, W1  = mg

When the lift descends with acceleration  a,

W_{2}=m(g-a)

\therefore\; \; \frac{W_{2}}{W_{1}}=\frac{m(g-a)}{mg}=\frac{9.8-5}{9.8}=\frac{4.8}{9.8}

or      W_{2}=W_{1}\times \frac{4.8}{9.8}=\frac{49\times 4.8}{9.8}=24N

Correct option is 1.

 

Option 1)

24 N

This is the correct option.

Option 2)

74 N

This is an incorrect option.

Option 3)

15 N

This is an incorrect option.

Option 4)

49 N

This is an incorrect option.

Explanation: