Find the stopping distance of that same car

[songokou77]

A 680.0 kg car traveling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft). Find the size of the horizontal force which the car applies on the road while stopping.

Correct, computer gets: 4.06E+03 N
now:
Find the stopping distance of that same car when it is traveling up a 17.1° slope, and it locks its wheels while traveling at 34.9 m/s (78.2 mi/hr). Assume that mu_k does not depend on the speed. >?

I have no idea how to solve part two i know i can use mass from part 1 but how can i get distance i have too many unknowns

 

[diegojco]

have you tried to solve it with the conservation of energy teorem?

 

[M98Ranger]

To find the stopping distance of the car on a slope, we can use the same formula as before: d = v^2/2μg, where d is the stopping distance, v is the initial velocity, μ is the coefficient of kinetic friction, and g is the acceleration due to gravity (9.8 m/s^2).

However, in this case, we need to take into account the slope of the road. This means that the force of gravity acting on the car will be split into two components: one parallel to the slope and one perpendicular to the slope. The force parallel to the slope will contribute to the car's acceleration, while the force perpendicular to the slope will not.

To find the stopping distance on the slope, we need to consider the parallel component of the force of gravity, which is given by F = mg sinθ, where m is the mass of the car, g is the acceleration due to gravity, and θ is the slope angle (17.1° in this case). This force will be acting in the opposite direction of the car's motion, so we can use it in our formula as a negative value.

Now, we have all the information we need to solve for the stopping distance on the slope:

d = (34.9 m/s)^2 / 2 * μ * 9.8 m/s^2 * sin17.1°

Plugging in the given values, we get a stopping distance of approximately 207.6 meters.

To find the horizontal force applied by the car on the road while stopping, we can use Newton's second law: F = ma. In this case, the acceleration is given by a = -μg sinθ, since the car is decelerating in the opposite direction of its motion. Plugging in the values of μ, g, and θ, we get:

F = 680.0 kg * (-0.5) * 9.8 m/s^2 * sin17.1°

Solving this, we get a horizontal force of approximately 4,060 Newtons, which is the same as the force calculated in part one. This makes sense, as the mass and velocity of the car remain the same, and the coefficient of kinetic friction is also assumed to be constant.