A car weighing 1000 kg and travelling at `30 m//s` stops at a distance of 50 m decelerating uniformly. What is the force exerted on it by the breaks ? What is the work done by the brakes?
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In order to calculate the force, we have to find out the acceleration (or rather retardation) first.
Now, Initial speed, u = 30 m/s
Final speed, v = 0 (The car stops)
Acceleration, a = ? (To be calculated)
And, Distance, s = 50 m
Now, we know that :
`v^(2) = u^(2) + 2as`
So, `(0)^(2) = (30)^(2) + 2 xx a xx 50`
100 a = – 900
`a = -(900)/(100)`
Thus, Acceleration, `a = – 9 m//s^(2)`
The force exerted by the brakes can now be calculated by using the formula:
`F = m xx a`
Here, Mass, `m = 1000 kg ” “` (Given)
And, Acceleration, `a = -9 m//s^(2)” “` (Calculated above)
So, Force, `F = 1000 xx (-9)`
F = – 9000 N
Thus, the force exerted by the brakes on the car is of 9000 newtons. The negative sign shows that it is a retarding force.
The work done by the brakes can be calculated by using the relation :
`W = F xx s`
Here, Force, F = 9000 N
Distance, s = 50 m
So, Work done, `W = 9000 xx 50 J`
= 450000 J
`= 4.5 xx 10^(5) J`
Thus, the work done by the brakes is `4.5 xx 10^(5)` joules.