Question

While standing at the top of a 2.5 meter ladder you decide to jump and gravity quickly pulls you down. If your mass is 78kg what is your kinetic energy as you are falling?
(A) $195\;{\rm{J}}$
(B) $1911\;{\rm{J}}$
(C) $3745.6\;{\rm{J}}$
(D) $31.2\;{\rm{J}}$

Answer 1

Whenever a conversion of energies occurs, then the conservation of energy can be used to calculate the value of converted energy. If there is no external loss of energy, then the law of energy conservation will act and the initial energy of the system will be equal to the final energy of the system.

Complete step by step answer:
Given,
The given mass is: $m = 78\;{\rm{kg}}$
The length of the ladder is:$l = 2.5\;{\rm{m}}$
The length of ladder from the bottom point to the top point will be the height from where I jumped.
So we can say that the value of height is:$h = 2.5\;{\rm{m}}$
We know that at the top of the ladder we have the total initial energy (${E_i}$) equal to the summation of initial kinetic energy ($K{E_1}$) and initial potential energy ($P{E_1}$).
Now we will write the expression for the total initial energy,
$\Rightarrow {E_i} = K{E_i} + P{E_i}\\\ \Rightarrow {E_i} = \dfrac{1}{2}m{v^2} + mgh..........{\rm{(1)}}$
Here, $m$ represents mass , $v$ is the velocity , $g$ is the acceleration due to gravity and $h$ is the height of the top of the ladder from the ground.
At the top of the ladder we have only potential energy and we have velocity equal to zero so the kinetic energy at the top of the ladder will be zero.
It is known that the acceleration due to the gravity is $9.81\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}$ .
Substitute all the values in the above expression,
$\Rightarrow {E_i} = \dfrac{1}{2} \times 78\;{\rm{kg}} \times {0^2} + 78\;{\rm{kg}} \times 9.81\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}} \times 2.5\;{\rm{m}}\\\ \Rightarrow {E_i} = 0 + 1911\\\ \Rightarrow {E_i} = 1911\;{\rm{J}}$
Now we will calculate the total final energy of the system after falling.

$$\Rightarrow {E_f} = K{E_f} + P{E_f}\\\ \Rightarrow {E_f} = K{E_f} + mg{h_f}$$

Here, $K{E_f}$is the final kinetic energy, $P{E_f}$is the final potential energy, and ${h_f}$ is the height of bottom of the ladder from the ground and that is zero.
Substitute all the values in the above expression,

$$\Rightarrow {E_f} = K{E_f} + 78 \times 9.81 \times 0\\\ \Rightarrow {E_f} = K{E_f}$$

Now we will use the law of energy conservation and calculate the final kinetic energy,
$\Rightarrow {E_i} = {E_f}\\\ \Rightarrow 1911\;{\rm{J}} = K{E_f}\\\ \Rightarrow K{E_f} = 1911\;{\rm{J}}$
Therefore, the final kinetic energy is $1911\;{\rm{J}}$ and the correct answer is option(B).

Note: In such a type of question the energy gained energy by the object after falling is the energy that is converted in from the energy due to its position into the energy due to its motion.