A driver of a 7.50 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.
I don't think I did it correctly, but here is my work...
(7.50x10^3N)(30m)=2.25x10^5J
Well, well, well, looks like someone's got some brake trouble! Let's see if we can sort it out with a little humor, shall we?
So, according to your calculations, the amount of work done by the brakes on the car is 2.25x10^5 Joules. Now, that's quite a lot of work for a little car, don't you think? I mean, it's not like the brakes are going to ask for a promotion or a raise or anything. They're just doing their job!
But seriously, let's break it down. When the car comes to a stop, the brakes are essentially converting the car's kinetic energy into heat energy through friction. So, yeah, the brakes are definitely working hard to save the day!
Now, whether your calculation is correct or not, that's a different story. Without knowing the car's velocity or the time it takes to stop, it's hard to say for sure. But hey, I appreciate the effort! Keep those wheels turning and those brakes pumping!
To find the work done by the brakes on the car, we need to use the work-energy principle. The work done by the brakes is equal to the change in kinetic energy of the car.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The kinetic energy of an object is given by the equation:
K = (1/2)mv^2
Where K is the kinetic energy, m is the mass of the object, and v is its velocity.
Since the car comes to a stop, its final velocity is 0. The initial velocity is not given in the problem statement, but since we know the weight of the car is 7.50 N, we can assume that it is initially at rest.
Therefore, the initial velocity is 0 m/s, and the final velocity is also 0 m/s.
The change in kinetic energy is then given by:
ΔK = Kf - Ki = (1/2)m(vf^2 - vi^2)
Since both final and initial velocities are 0, the change in kinetic energy is also 0.
Therefore, the work done by the brakes on the car is 0 J. This means no work needs to be done by the brakes to stop the car just in time.
To find the work done by the brakes on the car, you need to use the work-energy principle. The principle states that the work done on an object is equal to the change in its kinetic energy.
In this case, the car comes to a stop, so its final velocity is zero. The initial velocity is not given, but we can assume it is constant. Therefore, we can calculate the initial kinetic energy of the car.
The kinetic energy of an object can be calculated using the formula:
KE = (1/2) m v^2
Where:
KE = kinetic energy
m = mass of the object
v = velocity of the object
The mass of the car is not given, but we can exclude the weight of the driver, so we only need to consider the weight of the car itself. Assuming that the gravitational field strength is 9.8 m/s^2, we can calculate the mass of the car from its weight:
Weight = mass * gravitational field strength
7.50 N = mass * 9.8 m/s^2
mass = 7.50 N / 9.8 m/s^2
Now that we have the mass, we can calculate the initial kinetic energy of the car using the initial velocity.
Since the final velocity is zero and the acceleration is constant, we can use the following equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (zero in this case)
u = initial velocity
a = acceleration
s = distance
We know that the final velocity is zero, the initial velocity is unknown, the acceleration is also unknown, and the distance is given as 30 meters. We can solve for the initial velocity using this equation:
0 = u^2 + 2as
u^2 = -2as
u = sqrt(-2as)
Now, plug in the values: s = 30 m, a = acceleration (unknown), and solve for u.
In this case, the acceleration is unknown, but since the negative acceleration is constant, we can use the formula:
a = (v - u) / t
Where:
a = acceleration
v = final velocity (zero in this case)
u = initial velocity
t = time
Using the given values:
0 = u + 2a(30)
0 = u + 60a
u = -60a
Now, substitute u into u^2 = -2as and solve for a:
(-60a)^2 = -2a(30)
3600a^2 = -60a
Divide both sides by a:
3600a = -60
a = -60 / 3600
a = -0.0167 m/s^2 (rounded to four decimal places)
Now that we have the acceleration, we can solve for the initial velocity:
u = sqrt(-2as)
u = sqrt(-2 * (-0.0167) * 30)
u = sqrt(1)
u = 1 m/s
Now, we can calculate the initial kinetic energy:
KE = (1/2) m v^2
KE = (1/2) * mass * (1)^2
Plug in the value of mass we calculated earlier, and solve for KE.
Finally, the work done by the brakes on the car is equal to the change in kinetic energy:
Work = final kinetic energy - initial kinetic energy
Since the final velocity is zero, the final kinetic energy is zero:
Work = 0 J - initial kinetic energy
Substitute the value of initial kinetic energy and calculate the work done by the brakes.