Question 189: Laws of Motion

Question

The weight of man in a stationary lift is $w_1$ and when it is moving downwards with uniform acceleration ' a ' is $\mathrm{w}_2$. If the ratio $\mathrm{w}_1: \mathrm{w}_2=4: 3$, then the value of ' a ' is ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

Weight of a man when the lift is stationary, $\mathrm{W}_1=\mathrm{mg}$.
Weight when the lift is going down,
$$ \begin{array}{ll}  & W_2=m(g-a) \ 	herefore & \frac{W_1}{W_2}=\frac{m g}{m(g-a)}=\frac{4}{3} \,\,\,\,\,	ext { ....(Given) }\ 	herefore & \frac{g}{g-a}=\frac{4}{3} \ 	herefore & 3 g=4 g-4 a \ 	herefore & g=4 a \Rightarrow a=\frac{g}{4} \end{array} $$

Answer

$\frac{\mathrm{g}}{3}$

Answer

$\frac{\mathrm{g}}{4}$

Answer

$\frac{3}{4} \mathrm{~g}$

Answer

$\frac{4}{\mathrm{~g}}$

Laws of Motion

The weight of man in a stationary lift is w_1 and when it is moving downwards with uniform acceleration ' a ' is \mathrm{w}_2. If the ratio \mathrm{w}_1: \mathrm{w}_2=4: 3, then the value of ' a ' is ( \mathrm{g}= acceleration due to gravity)

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