Question

Derive an expression for kinetic energy.A car before stopping moving at speed of 5m/s with passanger of mass 1200kg and mass of car is 1180kg .What is the work done by break to stop car?​

Answer 1

${\huge{\pink{\underbrace{\overbrace{\mathbb{\blue{ANSWER:-}}}}}}}$

Derivation:-

We know,

Work done= kinetic energy

W.D.=K.E.

So,

K.E.= Fs   (F= force,s=displacement)

F=ma - - - - (i)

v²=u²+2as

⇒s=$\frac{v^{2}-u^{2}}{2a}$ - - - - (ii)

Lets substitute (i) and (ii) in (K.E=Fs)

K.E.=Fs

K.E=ma×$\frac{v^{2}-u^{2}}{2a}$

('a' can be canceled and u=0, starting from rest)

So,

K.E= $\frac{1}{2}mv^{2}$

Solution:-

Velocity(v)= 5m/s

Mass of passengers=1200kg

Mass of car=1180kg

Total mass= 1200+1180

                 =1380kg

Lets apply the formula here,

K.E.=$\frac{1}{2}mv^{2}$

      $\huge{\frac{1}{2}\times1380 \times 5^{2}$

      $\huge{690 \times 25}$

      $\huge\boxed{\red{=17250J}}$

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