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# A solid rectangular block of mass $200kg$ has the dimensions $l = 2m$, $b = 1m,$ $h = 0.5m.$ It lies on the horizontal floor on sides l and b. The minimum work needed to turn it so that it lies on the sides b and h is:A) $Zero$B) $1500J$C) $3000J$D) $2000J$

Last updated date: 20th Sep 2024
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Hint: This question can easily be solved, if we know the relation between work done and energy. Also, according to the question, the block has to be turned. So, we need to calculate the respective height to which it is turned. After this, finally we can conclude with the solution.

Complete step by step solution:
We know that, change in kinetic energy = change in potential energy = work done
This can also be represented as, $\Delta K.E = \Delta P.E = W$
Since, the block is in static position and after turning it again it will be in static position, so there will be change in its potential energy.
The mass of the rectangular block is given as $200kg$.
Initially, the block was laying on a horizontal floor on the sides l and b. After turning the block will lie on sides b and h. So, the new length will be at the center of the length, i.e. $\dfrac{l}{2}$ and the new height will be at centre of the height, i.e. $\dfrac{h}{2}$.
Now, using the relation of energy and work from step one we can write it as,
$W = \Delta P.E$
$\Rightarrow W = mg\dfrac{l}{2} - mg\dfrac{h}{2}$
$\Rightarrow W = mg\left( {\dfrac{l}{2} - \dfrac{h}{2}} \right)$
$\Rightarrow W = 200 \times 10\left( {\dfrac{2}{2} - \dfrac{{0.5}}{2}} \right)$
$\therefore W = 1500J$
Therefore, the required work done is $1500J$.

Hence, option (B), i.e. $1500J$ is the correct choice of the given question.

Note: Potential energy is the energy stored in a body while kinetic energy is the energy gained by a body due to its motion. Also, work done is the force required to displace a body. According to the question due to the force applied in turning the block the position is displaced and accordingly we need to take the respective height on which the mass of the block is concentrated.