A car (mass = 750 kg) is initially moving with velocity 30 m/s. The driver slowly presses on the brakes and bring the car to a stop, traveling 50 m in the process. Calculate the force of friction between the car and the ground that brings it to a stop. Draw a free-body diagram for the car and clearly explain the method behind your solution.
F * d = work done by friction = initial Ke = (1/2) m v^2
F * 50 = (1/2)(750)(900)
100 F = 750 * 900
F = 750 * 9
There are harder ways of course using
F = change of momentum/ time = 750*30/t
then
50 = (1/2)a t^2
where a = 30/t
so
100 = 30 t
t = 10/3 seconds to stop
a = 9 m/s^2 (about 1 g)
F = m a = 750* 9
To solve this problem, we can use Newton's Second Law, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m*a).
In this case, the car is initially moving with velocity 30 m/s and comes to a stop after traveling 50 m, so its final velocity is 0 m/s.
First, we need to calculate the acceleration of the car. We can use the kinematic equation:
v^2 = u^2 + 2*a*s
where:
v = final velocity (0 m/s)
u = initial velocity (30 m/s)
a = acceleration (unknown)
s = displacement (50 m)
Rearranging the equation to solve for acceleration (a), we get:
a = (v^2 - u^2) / (2*s)
= (0^2 - 30^2) / (2*50)
= (-900) / 100
= -9 m/s^2
The negative sign indicates that the acceleration is opposite to the initial velocity, which means it is deceleration in this case.
Now, we can calculate the net force acting on the car. The only force acting on the car that causes the deceleration is the force of friction between the car and the ground.
The force of friction (F_friction) can be calculated using Newton's Second Law:
F_friction = m * a
= 750 kg * (-9 m/s^2)
= -6750 N
Since the force of friction always opposes the motion, it is negative in this case.
Therefore, the force of friction between the car and the ground is 6750 N.
Free-body diagram:
A free-body diagram is a visual tool used to represent the forces acting on an object. In this case, the forces acting on the car are:
1. Weight (mg) acting downwards, with magnitude 750 kg * 9.8 m/s^2 = 7350 N
2. Normal force (N) acting upwards, which equals the weight and has a magnitude of 7350 N
3. Force of friction (F_friction) acting in the drag direction, with a magnitude of 6750 N
To draw the free-body diagram, you would represent the car as a simple rectangle and label the forces acting on it. The weight force will be vertically downwards, the normal force upwards, and the force of friction pointing opposite to the direction of motion.