Question

kaushik stopped a moving car of mass 1000 kg in 10 minutes if initial velocity of car was 2 meter s/second find the force applied by kaushik to stop the car​

Answer 1

Explanation:-

m = 1000 kg

t = 10 minutes , 600s

u = 2 m/s

v = 0 m/s

To find :-

The force applied to stop the car.

Solution:-

First we have to find the retardation produced by brakes.

• Using equation of motion:-

→$V = u + at$

→$0 = 2 + a \times 600$

→$-2 = 600a$

→$a = \dfrac{-2}{600}$

→$a = \dfrac{-1}{300}m/s^2$

Force,applied by brakes on the car is :-

• By using Newton 2nd law of motion.

→$F = ma$

→$F = 1000 \times \dfrac{-1}{300}$

→$F = \dfrac{-10}{3}$

→$F = -3.33N$

hence,

The applied force is -3.33N .

Answer 2

Answer:

F = - 3.33

Explanation:

Given :

Mass of car = 1000 kg

Velocity of car = 2 m / sec .

Time = 10 min

In term of second = 10 × 60 sec = 600 sec .

We have to find Force :

We know ,

F = m a

Also know a = v / t

Here acceleration is use to stopped .

Thus we will take negative to show that working for oppose [ in order to stop ] .

Putting values in formula we get

F = 1000 × - 2 / 600 N

F = - 10 × 2 / 6 N

F = - 20 / 6 N

F = - 10 / 3 N

F =  - 3.33 N

Hence - 3.33 N force applied by Kaushik to stop the car .

Extra points :

Force is product of mass and acceleration .

It is vector quality .

If the answer would be in negative that represents direction of force applied ,

But here id simply in magnitude .