Question

A truck of mass 5000 kg is moving with a velocity of 20 ms${}^{-1}$. To avoid an accident, the driver applies brakes to stop the truck in 5 seconds. Calculate the force applied by the brakes in order to stop the truck in the given time using work-energy theorem.

Answer 1

Given : Mass of truck = 5000 kg, Initial velocity = 20 ms${}^{-1}$, Final velocity = 0 ms${}^{-1}$, Time to stop the truck = 5 seconds.

Using the work energy theorem,
1/2 mv${}^2$ - 1/2 mu${}^2$ = F x s ......(1)

To calculate 's', we'll have to use the equations of motion.
Using 1st equation of motion,
v = u + at
=> 0 = 20 + a x 5
=> a = - 4 ms${}^{-2}$

Now, using 3rd equation of motion,
v${}^2$ - u${}^2$ = 2as
=> 0 - (5)${}^2$ = 2 x (-4) x s
=> s = 25/8 m

Now, substituting the value of 's' in equation (1),
=> 1/2 x 5000 x (0)${}^2$ - 1/2 x 5000 x (5)${}^2$ = F x 25/8
=> - 1/2 x 5000 x 25 = F x 25/8
=> F = - 20,000 J

Negative sign implies force and the displacement are in opposite directions.