) A car is traveling on a straight, horizontal road at 20 m/s when the driver slams on the brakes. The force of friction between the car and the road is 900 N. The mass of the car is 1800 kg.
(a) Make a free body diagram for the car on the set of axes provided.
(b) How far will the car go before stopping?
(b) f = m a ... a = f / m = 900 / 1800
a = .5 m/s^2
so it takes 40 s to stop ... 20 / .5
ave velocity is ... (20 + 0) / 2
distance = v * t
To answer this question, let's break it down into two parts:
(a) To make a free body diagram for the car, we need to consider the forces acting on the car. In this case, we have the force of friction between the car and the road, which is acting in the opposite direction of motion.
The free body diagram will include two forces:
1. The force of gravity (mg) acting vertically downward, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The force of friction (f) acting horizontally in the opposite direction of motion.
Here's an example of a free body diagram for the car:
^
|-------| |-------|
\ /
\_________/
Ffriction
|
|-------| |-------|
(b) To find out the distance the car will travel before stopping, we can use the concept of kinetic friction and the equations of motion.
The force of friction is given as 900 N. This force can be calculated using the equation: f = μ * N, where μ is the coefficient of friction and N is the normal force.
In the given question, the car is on a horizontal road, so the normal force is equal to the weight of the car, which is mg (mass * acceleration due to gravity).
Since the forces acting on the car are balanced (the net force is zero when the car comes to a stop), we can use the equation:
f = μ * N
900 N = μ * (m * g)
Solving for μ:
μ = f / (m * g)
μ = 900 N / (1800 kg * 9.8 m/s^2)
μ = 0.051
Now, we can use the equation of motion to find the distance traveled by the car before stopping. The equation is:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s because the car comes to a stop)
u = initial velocity (20 m/s, given in the question)
a = acceleration (which is equal to the friction force divided by the mass of the car)
s = distance traveled
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the values we have:
s = (0^2 - 20^2) / (2 * (-900 N / 1800 kg))
s = (0 - 400) / (-1)
s = 400 meters
Therefore, the car will travel 400 meters before stopping.