A truck and a car moving with the same kinetic energy are brought to rest by the application

of brakes which provide equal retarding forces $S$ . Both the truck and the car travel the ____ distance before coming to rest.

Question

A truck and a car moving with the same kinetic energy are brought to rest by the application

of brakes which provide equal retarding forces $S$ . Both the truck and the car travel the ____ distance before coming to rest.

Accepted Answer

A truck and a car moving with the same kinetic energy are brought to rest by the application
of brakes which provide equal retarding forces

S

. Both the truck and the car travel the different distance before coming to rest.
Let

M

and

V

be respectively the mass and velocity of the truck and m and v be the mass and
velocity of the car. Also, let us assume that they both have wheels made of the same material,
which has a friction coefficient

μ

with the asphalt. If both of them have the same kinetic
energy, we can say for sure that

T = \frac{1}{2} m v^{2}

= \frac{1}{2} M V^{2} - - - - (1)

And, consequently,

T = m v^{2}

= M V^{2}

T = \frac{v^{2}}{V^{2}}

= \frac{M}{m} - - - - - (2)

Now, we know that there’s only one force acting on each body on the horizontal direction —
since the weight and the floor’s normal reaction cancel each other. That would be the friction
force, which, for the car, is given by

F = μN

= μmg - - - - (3)

We should note that the work done by friction is the one responsible for converting the car’s
kinetic energy to other forms of energy — which in this case should be heated. As such, we can
say that

W = ΔT

\Rightarrow F \cdot d = \frac{1}{2} m v^{2}

—-(4)
where d is the distance travelled by car.
Plugging in (4) into that, we get

μmgd = \frac{1}{2} m \times v^{2}

\Rightarrow d = \frac{v^{2}}{2 g} - - - - (5)

In an analogous manner we find the distance D travelled by truck:

D = \frac{V^{2}}{2 g} - - - - - (6)

Dividing both equations (5) and (6), we have that

\frac{d}{D} = \frac{v^{2}}{V^{2}}

However, by (2), we can state that

\frac{d}{D} = \frac{v^{2}}{V^{2}}

Which is the general solution to your problem. By common sense, we should have the truck heavier than the car, which implies that

M > m

and therefore

\frac{M}{m} > 1

. That in turn, would assure that,

\frac{d}{D} > 1

, which is to say,

d > D

. Thus, we prove that the distance travelled by the truck should be shorter than the distance travelled by car if the previous is heavier than the
latter.