Question

A man of mass 60kg stands in a moving lift. What is the apparent weight when the lift moves (i)up with an acceleration of 4m/s²? (ii) down with an acceleration of 4m/s²? (iii) up with an uniform velocity of 4m/s?

Answer 1

Mass of man = 60 kg

Acceleration due to gravity = 9.8 $\(\frac{m}{s^2}\)$

(i) The apparent weight will increase when the lift moves up with an acceleration of 4 $\(\frac{m}{s^2}\)$.

Apparent weight = m(g+4)

= 60×(9.8 + 4)

= 60 × 13. 8

Apparent weight = 828 N

(ii) The apparent weight will decrease when the lift is accelerating downwards with acceleration 4 $\(\frac{m}{s^2}\)$.

Apparent weight = m × (g - 4)

= 60 × (9.8 - 4)

= 60 × 5.8

Apparent weight = 348 N

(iii) The apparent weight will be equal to true weight if the lift is moving with constant velocity.

Apparent weight = m × g

= 60 × 9.8

Apparent weight = 588 N

Answer 2

Answer:

When the lift moves upwards,the new force acts with the net force.

=mg +ma

=m(g+a)

=60(10+2)

=60(12)

F=720 N

When the body acts against the gravity, the new force tends to decrease the net force.

=m(g-a)

= 60(10-2)

=60(8)

F=480N

Hope this helped you