Question

A car of mass 1000kg moving at a speed 10 m/s is brought to rest in 40s . find the magnitude of the force applied​

Answer 1

Given:-

→ Mass of the car = 1000 kg

→ Initial velocity of the car = 10 m/s

→ Time taken to come to rest = 40s

To find:-

→ Magnitude of force applied.

Solution:-

In this case :-

• Final velocity of the car will be zero as it

finally comes to rest.

By Newton's 2nd law of motion, we know that :-

F = m(v - u)/t

Where :-

• F is the magnitude of force.

• m is mass of the body.

• v is the final velocity of the car.

• u is initial velocity of the car.

• t is time taken.

Substituting values, we get :-

=> F = 1000(0 - 10)/40

=> F = 25(-10)

=> F = -250 N

[Here, -ve sign represents retarding force]

Thus, magnitude of force applied is -250N or a retarding force of 250N .

Some Extra Information:-

2nd law of motion : The rate of change of momentum is directly proportional to the force applied and it takes place in the direction of force.

i.e. F ∝ m(v - u)/t

Answer 2

We have,

Mass of car = 1000 kg

Initial velocity = 10 m/s

Final velocity = 0 m/s

Time taken = 40 s

Now we know,

F = ma = m(∆v/∆t) =

⇒F = (1000 kg)[(0 m/s - 10 m/s)/(40 s)]

⇒F = (1000 kg)(- 0.25 m/s²)

⇒F = – 250 N (magnitude of force opposing it's motion).

∴ A force of 250 N acted on the car to stop it.