Question

2. A man weighing 55 kg on the earth is standing in a lift. Calculate the
apparent weight of the man if the lift is
(a) rising with an acceleration of 1 ms
(6) going down with an acceleration of 0.5 ms2
(c) falling freely under the action of gravity.​

Answer 1

w=55

g=9.8

a=1

55*(9.8+1)=594

ii)  a=0.5

t=55(9.8-0.5) =511.5N

iii)g=0 because freefalling

Answer 2

Given,

Mass of the man=55 Kg.

To find,

the apparent weight of the man if the lift is:

(a) rising with an acceleration of 1 $m/s^{2}$

(6) going down with an acceleration of 0.5 $m/s^{2}$

(c) falling freely under the action of gravity.

Solution:

• The concept that is to be used here is from the chapter "Laws of Motion".
• Normally the weight(actual weight) of the man will be,
• W=mg.
• W=55 x 10=550N.
• In the case of lift, when the lift goes upwards with an acceleration a, the apparent weight becomes more than the actual weight.
• The apparent weight then is given in the form of the normal reaction,N.
• N=m(g+a).
• where, N-normal reaction(apparent weight), m-mass of the body, g-acceleration due to gravity and a-acceleration of the lift.
• $g=10m/s^{2}$.
• When the lift goes downwards with an acceleration a, the apparent weight becomes less than the actual weight.
• N=m(g-a).
• When the lift is falling freely under gravity, the apparent weight is given as:
• N=mg.

(a)The apparent weight of the man when the lift is rising with an acceleration of 1 $m/s^{2}$ will be:

N=m(g+a)

N=55(10+1)

N=55 x 11 N

N=605N.

The apparent weight in this case is 605N.

(b)The apparent weight of the man when the lift is going  down with an acceleration of 0.5 $m/s^{2}$.

N=m(g-a)

N=55(10-0.5)

N=55 x 9.5

N=522.5N.

The apparent weight in this case is 522.5N.

(c))The apparent weight of the man when the lift is falling freely under the action of gravity.​

N=mg

N=55 x 10

N=550N.

The apparent weight in this case is 550N.

#SPJ2