Question

A 20 kg chair has 250 J of potential energy relative to the ground. If the chair is dropped from its rest position, what is its speed when it strikes the ground?

Answer 1

Given :

• Mass of the chair = 20 kg

• Potential energy= 250 J

To find :

The velocity of the chair when it strikes the ground.

Solution :

Let the velocity of the chair be v m/s.

We know that ,

Potential energy = Final kinetic energy i.e,

$\boxed{\bf{mgh = \dfrac{1}{2}mv^{2}}}$

Where :

• m = Mass of the body
• g = Acceleration due to gravity
• h = Height
• v = Velocity of the particle

Now , by using the above equation and substituting (given) values in it , we get :

$:\implies \bf{mgh = \dfrac{1}{2}mv^{2}}$

$:\implies \bf{250 = \dfrac{1}{2} \times 20 \times v^{2}}$

By multiplying 2 on both the sides , we get :

$:\implies \bf{250 \times 2 = \dfrac{1}{2} \times 20 \times v^{2} \times 2}$

$:\implies \bf{250 \times 2 = \dfrac{1}{\not{2}} \times 20 \times v^{2} \times \not{2}}$

$:\implies \bf{500 = 20 \times v^{2}}$

By dividing 20 on both the sides we get , we get :

$:\implies \bf{\dfrac{500}{20} = \dfrac{20 \times v^{2}}{20}}$

$:\implies \bf{25 = v^{2}}$

By square-rooting on both the sides, we get :

$:\implies \bf{25 = v^{2}}$

$:\implies \bf{\sqrt{25} = \sqrt{v^{2}}}$

$:\implies \bf{5 = v}$

$\boxed{\therefore \bf{v = 5\:ms^{-1}}}$

Hence, the velocity with which the chair will reach the ground is 5 m/s.

Answer 2

Ａｎｓｗｅｒ:–

$\overbrace{\underbrace{ \sf{ \dfrac{ \dfrac{Mass(m) = 20 kg }{ Potential \: energy(mgh) =250J} } { \dfrac{ Initial \: velocity(u) =0}{ Final ~velocity(v)= ?} }}}}$

$\star~ \boxed{\rm{ \red{mgh = \frac{1}{2}mv² }}}$

$→\rm 250 = \dfrac{1}{2} \times {v}^{2}$

$→ \rm 500 = {v}^{2}$

$→ \rm v = \sqrt{500}$

$\therefore~ \underline {\underline{\red{ \rm {v = 10\sqrt{5} \: m /s}}}}$