The apparent weight of a person inside a lift is w1 when lift moves up with a certain acceleration and is w2 when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

Question

The apparent weight of a person inside a lift is w1 when lift moves up with a certain acceleration and is w2 when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is (A)  (w1+w2/2)  (B)  (w1-w2/2)  (C)  2w1  (D)  2w2

Tardigrade Admin · Accepted Answer

When lift moves up with constant acceleration a, then

w_{1} - m g = ma \dots (i)

When lift moves down with constant acceleration a, then

m g - w_{2} = ma \dots (ii)

From Eqs. (i) and (ii), we get

w_{1} + w_{2} = 2 ma \dots (iii)

When lift moves up with constant speed,
its acceleration is zero
So,

w - m g = 0

or

w = m g \dots (i v)

From Eqs. (iii) and (iv), we get

w_{1} + w_{2} = 2 m g

or

w = \frac{w _{1} + w _{2}}{2}

Q. The apparent weight of a person inside a lift is $w_{1}$ when lift moves up with a certain acceleration and is $w_{2}$ when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

$\frac{w _{1} + w _{2}}{2}$

$\frac{w _{1} - w _{2}}{2}$

$2 w_{1}$

$2 w_{2}$

Q. The apparent weight of a person inside a lift is w1​ when lift moves up with a certain acceleration and is w2​ when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

2w1​+w2​​

2w1​−w2​​

2w1​

2w2​

Q. The apparent weight of a person inside a lift is $w_{1}$ when lift moves up with a certain acceleration and is $w_{2}$ when lift moves down with same acceleration. The weight of the person when lift moves up with constant speed is

$\frac{w _{1} + w _{2}}{2}$

$\frac{w _{1} - w _{2}}{2}$

$2 w_{1}$

$2 w_{2}$