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Net Force

A man standin...

Question

A man standing in a lift carrying a bag of $5 k g$ . If lift moves vertically downwards with accelerating $g / 2$ . Find tension force in handle of bag.

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Solution

Given,

$m = 5 k g$

$a = \frac{g}{2} m / s^{2}$

$g = 10 m / s^{2}$

From free body diagram,

$T = m a + m g$

$T = \frac{m g}{2} + m g = \frac{3 m g}{2}$

$T = \frac{3 \times 5 \times 10}{2}$

$T = 75 N$

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