Question

a car of mass 2500 kg with a velocity of 29m/s is stopped in 10 s by applying breaks so what is the retarding force
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Answer 1

Given:-

→Mass of the car = 2500kg

→Initial velocity of the car = 29m/s

→Time taken to stop = 10s

To find:-

→Retarding force

Solution:-

•Final velocity of the car will be zero as it finally stops after application of brakes.

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

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

Where:-

•F is force

•m is mass of the body

•v is final velocity of the body

•u is initial velocity of the body

•t is time taken

=>F = 2500(0-29)/10

=>F = 2500(-2.9)

=>F = -7250N

∵Magnitude of force is -ve

∴Magnitude of retarding force will be +ve

Thus,the magnitude of retarding force is 7250N.

Answer 2

Mass of the car (m) = 2500 kg

Initial Velocity of the car (u) = 29 m/s

As it was stopped finally , so

Final velocity of the car (v) = 0 m/s

Time taken to stop (t) = 10 s

___________________________

Force (F) : It is defined as mass times the acceleration

➳ F = ma

➳ SI Units : Newton  

Acceleration (a) : It is defined as rate of change in velocity

➳ a = (v - u) / t

➳ SI Units : m/s²

___________________________

➠ F = ma

➠ F = m (v - u) / t

➠ F = (2500) (0 - 29) / 10

➠ F = - 7250 N  $\pink{\bigstar}$

Note : - ve sign of force denotes retarding force

Retarding force = 7250 N