Question

A car is moving at 34.4 m/s and comes to a stop in 61.8 meters. Find the magnitude of the braking force applied to the car.

Answer 1

To find the magnitude of the braking force applied to the car, we can use the work-energy principle, which states that the work done by the braking force is equal to the change in kinetic energy of the car.

First, we need to calculate the initial kinetic energy of the car. The kinetic energy (KE) is given by the formula:

\[
KE = \frac{1}{2}mv^2
\]

where \( m \) is the mass of the car and \( v \) is its velocity. However, since we don't have the mass of the car, we will express everything in terms of \( m \).

The initial velocity \( v = 34.4 \, \text{m/s} \).

The final velocity when the car comes to a stop is:

\[
v_f = 0 \, \text{m/s}
\]

The initial kinetic energy then is:

\[
KE_i = \frac{1}{2}m(34.4)^2
\]

The final kinetic energy is:

\[
KE_f = \frac{1}{2}m(0)^2 = 0
\]

The work done by the braking force \( W \) is equal to the change in kinetic energy:

\[
W = KE_f - KE_i = 0 - \frac{1}{2}m(34.4)^2 = -\frac{1}{2}m(34.4)^2
\]

The work done by the force can also be expressed in terms of the force \( F \) and the distance \( d \):

\[
W = F \cdot d
\]

Since the car is coming to a stop, the work done is negative (because the force is acting in the opposite direction of the motion). Therefore:

\[
-F \cdot d = -\frac{1}{2}m(34.4)^2
\]

where \( d = 61.8 \, \text{m} \).

Thus, we have:

\[
F \cdot 61.8 = \frac{1}{2}m(34.4)^2
\]

Now we can solve for the braking force \( F \):

\[
F = \frac{\frac{1}{2}m(34.4)^2}{61.8}
\]

Next, we need to find \( \frac{m}{61.8} \) in order to isolate the force. To find the deceleration (a) caused by the force, we can use the kinematic equation:

\[
v_f^2 = v_i^2 + 2a d
\]

Substituting \( v_f = 0 \), \( v_i = 34.4 \), and \( d = 61.8 \):

\[
0 = (34.4)^2 + 2a(61.8)
\]
\[
(34.4)^2 = -2a(61.8)
\]
\[
a = -\frac{(34.4)^2}{2 \times 61.8}
\]

Calculating \( (34.4)^2 \):

\[
34.4^2 = 1183.36
\]

Now plug this value into the equation:

\[
a = -\frac{1183.36}{123.6}
\]
\[
a = -9.57 \, \text{m/s}^2
\]

Now we need to relate the braking force to acceleration:

\[
F = m \cdot a
\]
Substituting \( a \):

\[
F = m \cdot (-9.57)
\]

Now substituting \( m \) into the relationship we established before:

\[
F = -9.57m
\]

To find the exact force, we can recall that the force needed to stop was derived from \( -9.57m \):

Now we have the magnitude of the braking force in terms of mass. Since we established that force \( F = \frac{(34.4^2)}{2 \cdot 61.8} * m\):

Thus:

\[
|F| = |\frac{(34.4^2) \cdot m}{2 \cdot 61.8}|
\]
Calculating gives the magnitudes:

Therefore finalizing we can say:

Braking force equals mass times acceleration to resist it:

The magnitude of the braking force is:

\[
F ≈ 9.57m
\]

If we had the mass of the car, we could find the exact numerical value for the braking force. However, the magnitude will depend directly on the mass of the car multiplied by his deceleration.