Question

A car moving at 40 km/h is to be stopped by applying
brakes in the next 4.0 m. If the car weighs 2000 kg,
what average force must be applied on it ?​

Answer 1

Answer:

Initial Speed of the car, u = 40 km/h = 40 * 5/ 18 m/s = 11. 1 m/s

Final Speed of car, v = 0 m/s

Time, s = 4m

Weight of the car, m = 2000 kg

F = ma

By third equation of motion,

v^2 - u^2 = 2as

0 = ( 11.1) ^2 + 2*a * 4

0 = 123.21 + 8 a

a = - 123.21 / 8

a = - 15.41 m/s^2

F = ma

F = 2000 * ( - 15.41) N

F = - 30, 820 N

Answer 2

hay here s your ans >>>>>

Initial Speed of the car, u = 40 km/h = 40 * 5/ 18 m/s = 11. 1 m/s

Final Speed of car, v = 0 m/s

Time, s = 4m

Weight of the car, m = 2000 kg

F = ma

By third equation of motion,

v^2 - u^2 = 2as

0 = ( 11.1) ^2 + 2*a * 4

0 = 123.21 + 8 a

a = - 123.21 / 8

a = - 15.41 m/s^2

          .

       .      .

F = ma

F = 2000 * ( - 15.41) N

F = - 30, 820 N

Hope helped

dont think that i have coppied from google or the first answer i have the written copy if u want u can ask for..........................................

                                Sorry                                                    

                                    +                                                

                            Thankyou