Explanation
Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.
Thus,
$$1 joule = 1 newton \times 1 m\\$$
$$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$$
$$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$$
$$1 J = 10^{7} \ erg$$
Thus option D is correct.
Given W= work done
and mass = m
and vertical height= H
and gravitational force= g
W= mgh
We know that, the kinetic energy is
$$K.E=\dfrac{1}{2}m{{v}^{2}}....(I)$$
Checking all options given -
A) $$ m=3M $$ & $$ v=v $$
$$ K.E=\dfrac{1}{2}m{{v}^{2}} $$
$$ K.E=\dfrac{1}{2}\times 3M\times {{v}^{2}} $$
$$ K.E=\dfrac{3M{{v}^{2}}}{2} $$
B) $$ m=3M $$ & $$ v=2v $$
$$ K.E=\dfrac{1}{2}\times 3M\times 4{{v}^{2}} $$
$$ K.E=6M{{v}^{2}} $$
C) $$ m=2M $$ & $$ v=3v $$
$$ K.E=\dfrac{1}{2}\times 2M\times 9{{v}^{2}} $$
$$ K.E=9M{{v}^{2}} $$
D) $$ m=M $$ & $$ v=4v $$
$$ K.E=\dfrac{1}{2}\times M\times 16{{v}^{2}} $$
$$ K.E=8M{{v}^{2}} $$
Hence, the largest kinetic energy is $$9M{{v}^{2}}$$ when $$m=2M$$ and $$v=3v$$.
Total Energy = Kinetic Energy + Potential Energy
As object move upward is velocity will decrease due to deceleration by gravity.
So its kinetic energy =$$\dfrac{1}{2}m{{v}^{2}}$$ will decrease as it move upward.
But its potential energy =$$mgH$$ will increase as it move upward.
Hence, total energy will remain constant.
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