JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 24)
A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -
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$${{m{g^2}{t^2}} \over 8}$$
$${{3m{g^2}{t^2}} \over 8}$$
$$-$$ $${{m{g^2}{t^2}} \over 8}$$
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Explanation
N $$-$$ mg = $${{mg} \over 2}$$ $$ \Rightarrow $$ N = $${{3mg} \over 2}$$
The distance travelled by the system in time t is
S = ut + $${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$$
Now, work done
W = N.S = $$\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$$
$$ \Rightarrow $$ W = $${{3m{g^2}{t^2}} \over 8}$$
The distance travelled by the system in time t is
S = ut + $${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$$
Now, work done
W = N.S = $$\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$$
$$ \Rightarrow $$ W = $${{3m{g^2}{t^2}} \over 8}$$
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