Question

A 60Kg man is inside a lift which is moving up with an acceleration of 2.45 m/sec2. The apparent percentage change in his weight is

Answer 1

Given:

A 60Kg man is inside a lift which is moving up with an acceleration of 2.45 m/s².

To find:

Percentage change in his Weight.

Calculation:

Weight of the man in stationary lift be w1.

$\therefore \: w1 = mg$

$= > \: w1 = 60 \times 10$

$= > \: w1 = 600 \: N$

Weight of man in lift accelerating upwards be w2.

$\therefore \: w2 = m(g + 2.45)$

$= > \: w2 = 60 \times (10+ 2.45)$

$= > \: w2 = 60 \times ( 12.45)$

$= > \: w2 = 747 \: N$

Change in weight be ∆w.

∆w = 747 - 600 = 147 N

$\sf{\% \: change = \dfrac{147}{600} \times 100\%}$

$\sf{ = > \% \: change = 0.245 \times 100\%}$

$\sf{ = > \% \: change = 24.5 \%}$

So , final answer is :

$\boxed{ \sf{ \% \: change = 24.5 \%}}$

Answer 2

Explanation:

according to the question answer is given