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# A man of mass $50kg$ is standing in an elevator. If the elevator is moving up with an acceleration $\frac{g}{3}$ then what is the work one by normal reaction of the floor of the elevator on the man when elevator moves by a distance 12m? $\left(g=10m/{s}^{2}\right)$

A

8000 J

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B

6000 J

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C

4000 J

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D

2000 J

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Solution

## The correct option is A 8000 J Step 1: GivenMass of the man: $m=50kg$Distance moved by lift: $d=12m$Acceleration due to gravity is $g$.Acceleration of the lift: ${a}_{L}=-\frac{g}{3}$ (negative because the lift is going upwards, moving against earth's gravity)The resultant acceleration is $a$.Work done is $W$. Step 2: Formula used$F=m×a$$W=F×d$Step 3: Find the resultant acceleration by subtracting the acceleration of lift from acceleration due to gravity.$a={a}_{L}-g\phantom{\rule{0ex}{0ex}}=-\frac{g}{3}-g\phantom{\rule{0ex}{0ex}}=-\frac{g}{3}-g\phantom{\rule{0ex}{0ex}}=-\frac{4g}{3}$Step 4: Find the force using the formula and substitute the value of g and m (Take only the magnitude of acceleration as the negative sign only indicate direction and have no affect on value).$F=m×a\phantom{\rule{0ex}{0ex}}=50kg×\frac{4g}{3}m/{s}^{2}\phantom{\rule{0ex}{0ex}}=50kg×\frac{4×10}{3}m/{s}^{2}\phantom{\rule{0ex}{0ex}}=\frac{2000}{3}N$Step 5: Find the work done using the formula$W=F×d\phantom{\rule{0ex}{0ex}}=\frac{2000}{3}N×12m\phantom{\rule{0ex}{0ex}}=8000J$Hence, option (A) is correct.  Suggest Corrections  0      Similar questions  Explore more