16. A driver of a 7.50 x 103 N car passes a sign stating
“Bridge Out 30 Meters Ahead.” She slams on the brakes,
coming to a stop in 10.0 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
work = force * distance
30 m in 10 s ... average velocity is ... 3 m/s
final vel is zero , so initial is
... 3 m/s = (vi + vf) / 2 ... vi = 6 m/s
the work is equal to the initial kinetic energy of the car
K.E. = 1/2 m v^2 = 1/2 * (7.50E3 / 9.81) * 6^2 = ? Joules
To find the work done by the brakes on the car, we first need to find the force applied by the brakes.
Given:
Mass of the car, m = ?
Weight of the car, W = 7.50 x 10^3 N
Acceleration of the car, a = ?
Using Newton's second law of motion, we know that force is equal to mass times acceleration (F = ma). In this case, the force is the weight of the car provided (W = mg), so we can rewrite the equation as W = ma.
Here, we have the weight of the car (W) as 7.50 x 10^3 N. Rearranging the equation gives us:
a = W / m
To calculate the acceleration, we need to find the mass of the car. Unfortunately, the mass is not given in the problem. We can find the mass of the car by dividing its weight by the acceleration due to gravity (approximately 9.8 m/s^2):
m = W / g
Substituting the given weight, we get:
m = (7.50 x 10^3 N) / (9.8 m/s^2)
Now that we have the mass, we can determine the acceleration. Plug in the mass into the previous equation we derived:
a = (7.50 x 10^3 N) / m
Now we know the acceleration of the car caused by braking.
Next, we have to find the distance over which the car stops. The problem states that the car stops in 10.0 seconds, so we can use the formula of uniform acceleration to find the distance (d) traveled:
d = (1/2) * a * t^2
Using the known values, we get:
d = (1/2) * a * (10.0 s)^2
Now we have the distance over which the car stops.
Finally, to find the work done by the brakes, we can use the work-energy principle: work done is equal to the change in kinetic energy. In this case, the car starts with a certain amount of kinetic energy and comes to a stop, so the work done by the brakes is equal to the initial kinetic energy of the car. The initial kinetic energy can be calculated using the formula:
KE = (1/2) * m * v^2
Since the car comes to a stop, the final velocity (v) is 0. Therefore, the work done by the brakes is given by:
Work = (1/2) * m * (0)^2
Therefore, the work done by the brakes is 0 Joules, as no work is done when an object comes to a stop.
So, in this scenario, no work is required to stop the car just in time.