Question

a car of mass 1000 kg travelling at 10m/s is brought to rest over a distance of 20m. find the average braking force in newtons?​

Answer 1

Explanation:

u = 10 m/s

v = 0 m/s

S = 20 m

a = ?

Using the 3rd Equation of Motion,

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

$a = \frac{ {v}^{2} - {u}^{2} }{2s}$

$a = \frac{ - 100}{40} = - 2.5 \: m {s}^{ - 2}$

Here , - sign means that the car is retarding.

Now ,

$force = mass \times accleration$

$f = 1000 \times 2.5 = 2500 \: n$

Hence the average retarding force is 2500 N.

I hope it helps you. If you have any doubts, then don't hesitate to ask.

Answer 2

Answer :-

2500N

Explanation :-

Given :

Mass = 1000kg

Initial velocity of the car = 10m/s

Final velocity of the car = 0       [As car was going to stop]

Distance travelled = 20m

To Find :

Force = ?

Solution :

Average braking acceleration,

$\boxed{\sf{}v^2=u^2+2as}$

where,

v is the final velocity,

u is the initial velocity,

a is the acceleration and,

s is the distance.

Put their values and find “a”

$\implies \sf{}0=(10)^2+2\times a\times 20m$

$\implies \sf{}0=100+40\times a$

$\implies \sf{}-100=40\times a$

$\implies \sf{}-\dfrac{100}{40}=a$

$\therefore \sf{}a=2.5m/s^2$

Now,we know acceleration so let’s find out the force.

$\boxed{\sf{}F=ma}$

where,

F is the force,

m is the mass and,

a is the acceleration.

Put their values and find “F”

$\implies \sf{}F=1000\times 2.5$

$\implies \sf{}F=2500N$

Therefore,average braking force is equal to 2500N