Answer:
A car weighing 2500 kg and travelling at 15 m/s stops at a distance of 25 m deceleration uniformly. What is the force exerted on it by the brakes?
Ad by Ireland
Fall in love with the rich cultural heritage of Ireland.
Explore the hidden beauty and the rich cultural heritage of Ireland on your next trip.
Learn More
5 Answers
Naren Nagarajan
, Cambridge Examinations Officer at Kasiga School, Dehradun (2008-present)
Answered 3 years ago · Author has 1.1K answers and 1.9M answer views
(a) W= Fd ( energy = force x distance )
(b) KE = .5 m v*v ( Kinetic energy of the car )
KE =0.5 * 2500 * 225KE =281250 Joules
Using (a) , Reducing kinetic energy to zero over 25 m
281250 = F * 25
F = 281250 /25
F = 11250 N
To check or use a different method …..
v^2 = u^2 +2as ( Third equation of motion )
2as = v^2 -u^2 ( v = 0 and U = 15 )
as = -225/2
a = - 225/2 * 25
a= - 225/50
a= - 4.5 m sec ^ -2
F = ma
F = 2500 * - 4.5
F = - 11250 N ( the negative sign indicates that the force is opposite in direction to the motion of the car )
The answer is correct since both values for F are equal.
