A driver of a 7550 N car passes a sign stating "Bridge Out 25 Meters Ahead." She slams on the brakes, coming to a stop in 10 s. How much work must be done by the brakes on the car if it is to stop just in time? Neglect the weight of the driver, and assume that the negative acceleration of the car caused by the braking is constant.
Well, let's do a little bit of math here. The car comes to a stop, which means it goes from some initial speed to zero. We can use the formula:
V^2 = U^2 + 2as
where V is the final speed (zero in this case), U is the initial speed, a is the acceleration, and s is the distance covered.
Since the car stops just in time, we can assume that the distance covered (s) is equal to the distance to the bridge, which is 25 meters.
Now, let's rearrange the formula:
2as = V^2 - U^2
The initial speed (U) is not given, so we'll assume it's some positive value. The final speed (V) is zero.
2as = 0^2 - U^2
Simplifying, we get:
2as = - U^2
We know that the acceleration (a) is caused by the braking force. The work done by the brakes is equal to the force applied multiplied by the distance over which it acts:
Work = force × distance
In this case, the force is equal to the mass of the car multiplied by the acceleration:
Work = mass × acceleration × distance
Since the acceleration is negative (braking force), we can rewrite it as:
Work = - mass × |acceleration| × distance
Now substituting the formula for acceleration (a) from before:
Work = - mass × |- U^2 / (2s)| × s
Simplifying further:
Work = mass × U^2 / 2
So, the amount of work that must be done by the brakes on the car is equal to half the mass of the car multiplied by the square of its initial speed. And I must say, that's a lot of work for the poor brakes!
To find the work done by the brakes, we need to determine the net force acting on the car and the distance over which the force is applied.
First, let's find the net force acting on the car. We can use Newton's second law of motion, which states that the net force (F_net) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a):
F_net = m * a
In this case, the net force is the force applied by the brakes, and the acceleration is opposite in direction to the initial motion of the car, so we have:
F_net = -m * a
Next, let's find the acceleration. We know that the car starts from an initial velocity of zero and comes to a stop in 10 s. The acceleration can be calculated using the equation:
a = (final velocity - initial velocity) / time
Since the car comes to a stop, the final velocity is zero. Therefore, the acceleration is given by:
a = (0 - 0) / 10 s = 0 m/s^2
Now, we have the acceleration. Let's find the net force by substituting the values into the equation:
F_net = -m * a = -m * 0 = 0 N
Since the net force is zero, no work is done by the brakes. This is because work is equal to the force applied multiplied by the distance over which the force is applied (W = F * d), and if the force is zero, the work done is also zero.
Therefore, the amount of work that must be done by the brakes for the car to stop just in time is zero.