Question

24. The weight of a man in a lif moving upwards with an
acceleration 'a' is 600 N. When the lif moves downwards
with the same acceleration his weight is found to be 360 N.
The real weight of the man is
(b) 600 N
(c) 480 N
(d) 700 N
(a) 380 N​

Answer 1

Answer:

Given:

When the lift is moving up with acceleration "a" upwards, the weight of man is 600 N.

Again, when the lift goes down with same acceleration, the weight of the man becomes 360 N

To find:

Actual weight of the man. [ when the lift I stationary]

Concept:

Whenever the man is standing in a non-inertial reference frame (i.e accelerating lift) , he experiences PSEUDO-FORCE in the direction opposite to the direction of acceleration.

Calculation:

Let PSEUDO-FORCE be F

When lift is going up:

mg + F = 600 .....(1)

When lift is going down:

mg - F = 360 ......(2)

Solving the 2 Equations:

(1) + (2) = 600 + 360

=> 2mg = 960

=> mg = 480 N

So, actual weight of the man is 480 N.

Answer 2

Answer:

$\bold\red{c)480\:N}$

_______________________

Solution:-

Let the force be " W "

Firstly,

When the lift was going upwards,

Mg + W = 600 N...........(eq.1)

Secondly,

When the lift moves downwards,

Mg - W = 360 N............(eq.2)

From the above 2 equations,

2Mg = 960 N

=> Mg = 960/2 N

=> Mg = 480 N

So,real weight of the man is 480 N

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