Question

A car of mass 100 kg is moving with a speed of 20m/s. Calculate the kinetic energy.​

Answer 1

Answer:

F=m.a By Newton's second law

-500=100*a

i.e. a= -5 m/s

V^2 - U^2= 2 a s

0–20^2= 2.-5.s

-400= -10s

Or s= 40 meters.

The same can also be achieved if we consider,

Force is the rate of change of momentum according to 2nd law of motion

F=dP/dt i.e. -500 = P2-P1/T2-T1

Where P2 is initial momentum

Now if the car produces a constant force through the brakes and stops after t2 seconds from the application of brakes(I e.t1=0)

Then we have

-500= mv2-mv1/T2-0

-500= 100(0–20)/t2 since m=100kg and V2=0 rest and v1 = 20m/s

Simplifying this we get

-5= -20/t2

Or t2= 4

Hence the brakes would stop the car in 4 seconds.

Now acceleration or deceleration is rate of change of velocity,

So a= v2-v1/t2-t1 i.e. a= 0–20/4=-5m/s^2

V^2 - U^2= 2 a s

0–20^2= 2.-5.s

-400= -10s

Or s= 40 meters.

Hence the car would stop in 40meters.

This one is a longer way to arrive at the same answer.

Answer 2

In context to question asked,

We have to determine the kinetic energy.

As per question,

It is given that,

Mass of car = 100 Kg

Speed of car = 20 m/sec.

As we know that,

$KE =\frac{1}{2}mv^2$

So, putting the value of m and v,

We will get,

$KE = \frac{1}{2}\times 100 \times 20^2\\=>KE = 20000\:Kgm^{-1}sec^{-1}$