Question

A 3000 kg truck moving at a speed 72 km per hr stops after covering some distance ,the force applied by brakes is 24000 N . compute the distance covered and work done by this force ?

Answer 1

Given ;
m = 3000 kg
F = 24000 M
v = 0
u = 72 km/hr
= 72×(1000/3600) m/s
= 20 m/s
____________________
F = ma
-24000 = 3000 × a
a = -8 m/s^2
___________________
v^2 - u^2 = 2aS
0 - (20×20) = 2×(-8)×S
S = 400/16
S = 25 m
___________________
Work done = Change in Kinetic energy
= m(v-u)
= 3000(0-20)
= - 60000 J
___________________

Answer 2

Here,
v = 0
F = 24000N
u = 72km/hr = 72 x 5/18 = 20 m/s
m = 3000kg

F = ma
24000 = 3000 x a
a = 8m/s^2
But it is a case of retardation.
So acceleration = -8m/s^2

v^2 - u^2 = 2as
0 - 400 = -16s
s = 400/16 = 25 m

Now work done = Change in kinetic energy
m(1/2v^2 - 1/2u^2) = 3000( 1/2 (0)^2 - 1/2 (20)^2) = 3000(20) = 60000 J = 60 kJ