Question:

You live at the top of a steep (a slope of ???? degrees above the horizontal) hill and must park your 2200 kg car on the street at night.

a) You unwisely leave your car out of gear one night and your handbrake fails.

Assuming no significant frictional forces are acting on the car, how quickly will it accelerate down the hill?

b) You resolve to always leave your car in gear when parked on a slope. If the rolling frictional force caused by leaving the drive--?train connected to the wheels is 5000 N, at what rate will your car accelerate down the hill if the handbrake fails again?

*It was written on my homeowork that the answer for a=2.5m/s^2 and b=0.3m/s^2 but I don't know how they got that answer.

a) I don't know how they got that answer either.

depends on ????
mg sin???? = ma
b) mg sin????- Ff = ma

*It was written on my homeowork that the answer for a=2.5m/s^2 and b=0.3m/s^2 but I don't know how they got that answer.

the sin of the angle is 15 degrees. I just had the exact same homework problem and got the correct value for acceleration by doing what he said above.

adding to what I just posted, so the Fnet is = Force of the rolling car (mgsin(15))-Force of friction.

So Fnet=5580N-5000N=580N
And Fnet=ma
So acceleration=Fnet/mass
580N/2200kg= 0.264m/s^2

Why is it mgSin15 and not mgtan15 please?

mgsin15=ma

2200x10xsin15=5694=ma
5694=2200a
a=2.5m/s

To calculate the acceleration of the car in each scenario, we need to understand the forces acting on the car.

Firstly, we need to determine the angle of the slope. The question does not provide this information, so we cannot provide an exact answer. However, we can proceed assuming a specific angle for the sake of explanation.

a) In scenario a, the car is not in gear, and the handbrake fails. Since no significant frictional forces are acting on the car, the only force acting on it is the force due to gravity, which is directed down the slope. The magnitude of this force depends on the angle of the slope.

To find the acceleration, we can use the formula:

acceleration = force / mass

The force due to gravity can be calculated using the formula:

force = mass * gravitational acceleration * sin(angle)

Here, the gravitational acceleration is approximately 9.8 m/s^2.

If the angle of the slope is, for example, 10 degrees (which is relatively steep), we can substitute the values into the formula:

force = 2200 kg * 9.8 m/s^2 * sin(10 degrees) ≈ 359 N

Now, we can find the acceleration:

acceleration = 359 N / 2200 kg ≈ 0.1636 m/s^2

Therefore, if the slope is 10 degrees, the car will accelerate down the hill at approximately 0.164 m/s^2.

b) In scenario b, the car is left in gear, and the rolling frictional force caused by leaving the drive-train connected to the wheels is 5000 N. In this case, there are two forces acting on the car: the force due to gravity and the rolling frictional force.

To find the net force, we need to subtract the rolling friction force from the force due to gravity:

net force = force due to gravity - rolling frictional force

Assuming the same angle of the slope, we can calculate the force due to gravity as before:

force due to gravity ≈ 359 N

Substituting both values, we find the net force:

net force = 359 N - 5000 N ≈ -4641 N

Here, the negative sign indicates that the net force is acting in the opposite direction of the force due to gravity.

Now, we can find the acceleration:

acceleration = net force / mass

acceleration = -4641 N / 2200 kg ≈ -2.1091 m/s^2

Therefore, if the slope is 10 degrees and the rolling frictional force is 5000 N, the car will accelerate down the hill at approximately -2.109 m/s^2.

Please note that the values provided in your homework might be based on a different assumed slope angle or other variables. Adjusting these variables will change the final answers.