A car is speeding along a level road at 25 m/s. The brakes abruptly bring the500 kg car to a speed of 3 m/s over a distance of 5 m. Determine the force applied by

brakes and the work done by the car on the brakes.

Yes you can calculate the work done by calculating the change in the kinetic energy. The change in kinetic energy would simply be the difference between the Final and the initial kinetic energies. [find out yourself] 
Feel free to ask if you have any doubts.

  • -4

initial velocity(u)= 25m/s

final velocity(v)= 3m/s

mass(m)= 500kg

displacement (s)= 5m

acceleration(a)= v2-u2/2s

=(3)2-(25)2/2x5

=9-625/10

=(-61.6m/s2)

force(f)= ma

=500x(-61.6)

=(-30800)N

work done = fs

=(-30800)x5

=(-154000)J

  • 3

Mass of the car, m = 500 kg

Initial speed, u = 25 m/s

Final speed, v = 3 m/s

Displacement, S =5 m

From the third equation of motion

v2 u2=2aS

⇒ (3)2 (25)2 =2*a*5

⇒ a = -616/10

m/s = 61.6 m/s2

Force applied by the brakes,

F = ma

= 500 kg × 61.6 m/s2

= 30800 N

The work done by the car on the brakes,

W=FS

= 30800 N*5 m

= 154000 J

  • 2

dear meritnation

can we solve this question the the formula---------work=final kinetic energy-initial kinetic energy? if yes how ,if not why

  • 1

dear meritnation

can we solve this question by the formula---------work=final kinetic energy - initial kinetic energy?

if yes how ,if not why

  • 1
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