Question

car of mass 500 Kg. is moving with the velocity 36 Km/Hr. and brought to rest in 5 seconds by applying brakes to it. Calculate a) average force applied by brakes on car b) distance traveled by car before come to rest.​

Answer 1

Given:-

→ Mass of the car = 500kg

→ Initial velocity of the car = 36km/h

→ Final velocity of the car = 0

[ as it finally comes to rest ]

→ Time taken = 5s

To find:-

→ Average force applied

→ Distance travelled

Solution:-

Firstly, let's convert the initial velocity of the car from km/h to m/s.

=> 1km/h = 5/18m/s

=> 36km/h = 5/18×36

=> 10m/s

Now, let's calculate the acceleration of the car using 1st equation of motion:-

=> v = u + at

=> 0 = 10 + a(5)

=> -10 = 5a

=> a = -10/5

=> a = -2m/s²

[ Here, -ve sign shows retardation]

According to Newton's 2nd law of motion:-

=> F = ma

=> F = 500(-2)

=> F = -1000 N

[ Here,-ve sign represents retarding force]

Now, let's calculate the distance travelled by using the 3rd equation of motion:-

=> v² - u² = 2as

=> 0 - (10)² = 2(-2)s

=> -100 = -4s

=> s = -100/-4

=> s = 25m

Thus:-

• Average force applied by the brakes of

the car is a retarding force of 1000N.

• Distance travelled by the car before it

came to rest is 25m.

Answer 2

Given :

Mass of the car, m = 500kg

Initial velocity of the car, u = 36km/h

Final velocity of the car, v = 0m/s

Time taken, t = 5s

To find :

(a) Average force applied

(b) Distance travelled

Solution :

First of all we have to convert km/hr to m/s

1km/h = 5/18m/s

36km/h = 36 × 5/18 = 10m/s

(a) As we are provided with with initial Velocity, final velocity and time we can use 1st equation of motion,

➠ v = u + at

➠ 0 = 10 + a(5)

➠ -10 = 5a

➠ a = -10/5

➠ a = -2m/s²

[negative accerlation is called retardation]

now, using Newton's 2nd law of motion .i.e.,

The rate of change of linear Momentum of a body with time is proportional to the net external force acting on it.

In order words,

➠ F = ma

➠ F = 500(-2)

➠ F = -1000 N

➠ |F| = 1000N.

thus,Average force applied by brakes on of 1000N.

Remember !

SI unit of froce is Newton (N).

(b) Now, using 3rd equation of motion .i.e.,

➠ v² - u² = 2as

➠ 0 - (10)² = 2(-2)s

➠ -100 = -4s

➠ s = -100/-4

➠ s = 25m${.}$

thus, Distance travelled by the car before it came to rest is 25m.