Question

A 1000kg car is travelling at the rate of 108km/h when the breaks are applied .If the car stops in 10m, find the stopping force in newton.​

Answer 1

Answer:

mass of car=1000kg

u=108km/h or 30m/s

v=0m/s

s=10m

v^2=u^2+2as

0^2=30^2+2*a*10

-900=20a

a=-45m/s^2

F=ma

F=1000*45

F=45000N

Answer 2

given:

mass=1000 kg

initial velocity, u=108 km/h=108×5/18 m/s=30m/s

distance=10 m

find :

stopping force=?

force = mass ×acceleration

f= 1000× a

finding acceleration,

v²=u²+2as( u is initial velocity)(v is final velocity =0)

a= v²-u²/2s= 0- 30²/2(10)

a= -900/20= -45 m/s² (-ve sign shows that deccelaration)

F =1000×(-45) = - 45000 N

(-ve sign shows that force is acting opposite to the motion)

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