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.
To find the work done by the brakes, we need to use the work-energy theorem. According to the theorem, the work done on an object is equal to the change in its kinetic energy. In this case, the car is initially moving with a certain velocity and comes to a stop, so its initial kinetic energy will be converted to zero.
The change in kinetic energy (∆KE) is given by the formula:
∆KE = KE_final - KE_initial
Since the car comes to a stop, its final kinetic energy (KE_final) would be zero. Therefore:
∆KE = 0 - KE_initial
= -KE_initial
Now, let's calculate the initial kinetic energy. The kinetic energy of an object is given by the formula:
KE = 0.5 * mass * velocity^2
To find the mass of the car, we can use the equation:
force = mass * acceleration
In this case, the force is the weight of the car, which is given as 7550 N. The acceleration is the negative acceleration caused by braking. As the negative acceleration is constant, we can calculate it by dividing the change in velocity (∆v) by the time taken (t) for the car to come to a stop.
∆v = 0 - initial velocity
= -initial velocity
acceleration = ∆v / t
= (-initial velocity) / t
Substituting the force and acceleration values, we can solve for mass:
7550 N = mass * ((-initial velocity) / t)
Now, let's substitute the expression for mass into the formula for kinetic energy:
KE_initial = 0.5 * (7550 N) * initial velocity^2 / t^2
Finally, plugging this value of KE_initial into the equation for ∆KE, we get:
∆KE = -0.5 * (7550 N) * initial velocity^2 / t^2
The negative sign in front of ∆KE indicates that the work done by the brakes is in the opposite direction to the initial motion of the car. Thus, the magnitude of the work done by the brakes is:
|W| = |-∆KE|
= 0.5 * (7550 N) * initial velocity^2 / t^2
This is the amount of work that must be done by the brakes on the car to stop it just in time.