Question

Q6.  Calculate the force exerted by the brakes on the motorcar which is moving with a velocity of 72 km/h and takes 5 s to stop after the application of brakes. Mass of the motorcar along with the passengers is 800 kg.​

Answer 1

Given

→Initial velocity of the motorcar = 72km/h

→Mass of the motorcar = 800kg

→Time taken to stop = 5s

To find:-

→Force exerted by the brakes of the

motorcar

Solution:-

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

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

=> 72km/h = 5/18×72

=> 20m/s

•Final velocity of the motorcar will be zero, as it finally comes to rest after application of brakes.

By using Newton's 2nd law of motion, we get:-

=> F = m(v-u)/t

Where:-

• F is the force

• m is mass if the body

• v is final velocity of the body

• u is initial velocity of the body

• t is time taken

=> F = 800(0-20)/5

=> F = 800(-20)/5

=> F = 800(-4)

=> F = -3200 N

[Here,-ve sign represents retarding force]

Thus, force exerted by the brakes of the motorcar is -3200N, or a retarding force of 3200N.

Answer 2

Given :-

☄ lnitial velocity, u = 72km/hr = 20m/s

☄ finial Velocity, v = 0m/s

☄ time taken, t = 5s.

☄ Mass, m = 800kg.

To find :-

the force exerted by the brakes on the motorcar

Solution :-

Using equation of motion .i.e.,

➠ v = u + at

➠ 0 = 20 + (a)(5)

➠ a = -20/5

➠ a = -4m/s²

|| -ve sign detects that the body is Retarding ||

we know that,

➠ F = ma

➠ F = (800)(4)

➠ F = 3200N

thus, the force exerted by the brakes on the motorcar is 3200N.

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