A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s2   the reading of the spring balance will be

  • Option 1)

    24 N

  • Option 2)

    74 N

  • Option 3)

    15 N

  • Option 4)

    49 N

 

Answers (1)

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, W= 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.

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