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A car weighing $1000 kg$ and travelling at $30 m / s$ stops at a distance of $50 m$ decelerating uniformly. What is the force exerted on it by the brakes?

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+ 15000 N

- 9000 N

- 15000 N

+ 10200 N

answer is B.

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Detailed Solution

The force exerted by brakes is

- 9000 N

.
Given data:
Mass of car,

m = 1000 kg

Initial speed of the car,

u = 30 m / s

The car stopped after the break then the final speed of the car,

v = 0 m / s

Displacement during the apply break,

s = 50 m

Step 1:
Find the acceleration of the car.
As we know, from the equation of motion.

v^{2} = u^{2} + 2 as

Where

v

is the final speed,

u

is the initial speed,

s

is the displacement and

a

is the acceleration.

0^{2} = 30^{2} + 2 \times a \times 50

100 a = - 900

a = - \frac{900}{100}

a = - 9 m / s^{2}

…..(i)
Thus, Acceleration,

a = - 9 m / s^{2}

Step 2:
Find the force exerted by the brakes.
Formula used:

F = ma

Where

F

is the force,

m

is the mass of the object and

a

is the acceleration.
Now,

F = 1000 \times (- 9)

[From equation (i)]

F = 1000 \times (- 9)

F = - 9000 N

…… (ii)
The negative sign shows that it is a retarding force.

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