Question

An object of mass 5 kg is raised 10 m above the ground. If it is allowed to fall, its kinetic energy when it is reached $\frac{1}{4}th$ way down is [Take acceleration due to gravity as $10m/s^2$]

• A. $125J$
• B. $500J$
• C. $375J$
• D. $275J$

Answer 1

The correct option is A $125J$

Potential energy of body at height h from the ground, i.e., at point B, P.E. = mgh

$\Rightarrow P.E.=5\times 10\times 10=500J$





Since the object is dropped from rest, its initial velocity, u is 0.

Kinetic energy at point $B=\frac{1}{2}m{u}^2(K.E.{)}_B=\frac{1}{2}\times 10\times 0^2=0$

Total Mechanical energy, $(M.E.{)}_B=(P.E.{)}_B+(K.E.{)}_B\Rightarrow (M.E.{)}_B=500J+0=500J\dots ..\left(i\right)$

Potential energy at point C $=mg\times \left(AC\right)=mg\left(\frac{3h}{4}\right)=\frac{5\times 10\times 3\times 10}{4}=375J$

Let kinetic energy at point C be $(K.E.{)}_C.$

$\therefore$ Mechanical energy at point $C,(M.E.{)}_C=375J+(K.E.{)}_C\dots ..\left(ii\right)$

Using law of conservation of energy:

Mechanical energy at point B = Mechanical energy at point C
$(M.E.{)}_B=(M.E.{)}_C\Rightarrow 500J=375J+(K.E.{)}_C\Rightarrow (K.E.{)}_C=(500–375)J=125J$

Hence, the correct answer is option (a).