Question

26. A child weight 20kg standing inside a lift
(i) If the lift is moving down with uniform acceleration of
5m/s2, what will be the apparent weight of the child? (2)
(ii) ) If the lift is moving up with uniform acceleration of 5m/s2,
what will be the apparent weight of the child (2)
(iii) If the lift cable breaks what will be the apparent
Weight of the child (1)​

Answer 1

Given :

Mass of the child = 20 kg

Acceleration of the lift = 5 m/s²

To Find :

The apparent weight of the child

Solution :

Case 1:

• If the lift is moving upward the acceleration of the lift is equal to (g-a) as the lift travels against accelerattion due to gravity.

       $W=m(g-a)$

       $W=20(9.8-5)$

       $W=96N$

The apparent weight of the child is 96N.

Case 2:

• If the lift is moving downward the acceleration of the lift is equal to (g+a) as the lift travels along accelerattion due to gravity.

       $W=m(g+a)$

       $W=20(9.8+5)$

       $W=296N$

The apparent weight of the child is 296N.

Case 3:

• If the cable breaks down, the lift travels with acceleration equal to acceleration due to gravity.

       $W=mg$

       $W=20(9.8)$

       $W=196N$

The apparent weight of the child is 196N.

Answer 2

Answer:

Explanation:

Given :

Mass of the child = 20 kg

Acceleration of the lift = 5 m/s²

To Find :

The apparent weight of the child

Solution :

Case 1:

If the lift is moving DOWNWARD the acceleration of the lift is equal to (g-a) as the lift travels against accelerattion due to gravity.

     

     

     

The apparent weight of the child is 96N.

Case 2:

If the lift is moving UPWARD the acceleration of the lift is equal to (g+a) as the lift travels along accelerattion due to gravity.

     

     

     

The apparent weight of the child is 296N.

Case 3:

If the cable breaks down, the lift travels with acceleration equal to acceleration due to gravity.

     

     

     

The apparent weight of the child is 0N.