# Calculating Cable Car's Speed on Hill Descent - 4122N Braking Force

• bluebear19
In summary, the cable car needs a braking force of 4122N to descend at constant speed. If the brakes fail and the cable car starts its downward journey, the runaway car's speed at the bottom of the hill will be 28.9m/s, assuming negligible rolling friction and using conservation of energy.
bluebear19

## Homework Statement

The 2100kg cable car shown in the figure descends a 200-m-high hill. In addition to its brakes, the cable car controls its speed by pulling an 1840kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

How much braking force does the cable car need to descend at constant speed?

One day the brakes fail just as the cable car leaves the top on its downward journey. What is the runaway car's speed at the bottom of the hill?

F = ma

## The Attempt at a Solution

I already found the breaking force = 4122N
but my answers for the second question was always wrong

#### Attachments

• knight_Figure_08_39[1].jpg
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Last edited:

Use conservation of energy.

$$U_g = mg\Delta h$$

$$E_k = \frac{1}{2}mv^2$$

Last edited:

So Ug = 4122*200 = 824600?
what does Ug or Ek stand for?

I'm really worried because I have to submit my answer online and if I get it wrong, its not like I can change it. thanks so much for helping!

bluebear19 said:
So Ug = 4122*200 = 824600?
No.
what does Ug or Ek stand for?
Gravitational potential energy and kinetic energy.

When considering conservation of energy, be sure to consider both the car and the counterweight.

so does Ek = 824600 = .5mv^2
then 824600 = .5(4123/9.8)v^2
and v = 62.6m/s?
i don't know if I'm doing this right at all

There are at least two (equivalent) ways to attack this problem. You can use the result from part one to find the net force on the system and then its acceleration. Then you can find the final speed using kinematics.

Or you can use conservation of energy: ΔKE + ΔPE = 0. Start by finding the change in PE of the system. Hint: The car goes down while the counterweight goes up.

so if i use the first way and my results from the first part which the net force = 4120N. That's the right answer

then I do F = ma
4120 = (2100sin30 - 1840sin20)a
a = 9.8m/s^2
isnt that the same as g?

so if that's right then I use : V_s^2 = V_0^2 + 2a(delta x)
delta x: sin30 = 200/x
x = 400m
V_s^2 = 0 + 2(9.8)(400)
v = 88.5m/s

i did this and the answer was wrong, can you please tell me what I did wrong, thanks so you much!

bluebear19 said:
so if i use the first way and my results from the first part which the net force = 4120N. That's the right answer
OK. If the breaking force is removed, then 4120 N is the net force.

then I do F = ma
4120 = (2100sin30 - 1840sin20)a
m is the mass of the system. Just add the masses.
a = 9.8m/s^2
isnt that the same as g?
Getting a = g should tip you off that something's wrong.

so I don't have to take into account the angles of the slopes at all?

so then 4120 = (2100+?? or - 1840) because isn't the counterweight slowing the car down? so should I add the masses

and then do I use the same kinematics that I used before??

bluebear19 said:
so I don't have to take into account the angles of the slopes at all?
Not when calculating the total mass.

so then 4120 = (2100+?? or - 1840) because isn't the counterweight slowing the car down? so should I add the masses
The total mass is just the sum of both masses. Just add them.

and then do I use the same kinematics that I used before??
Yes.

4120 = (2100 + 1840)a
a = 1.04 m/s^2

V_f^2 = 0 + 2(1.04)(400)
V_f = 28.9m/s

so the 400 m is right?
and would the final velocity be negative since its going downhill?

nevermind I got it thank you sooooo much

## 1. How is the cable car's speed calculated on a hill descent?

The cable car's speed on a hill descent is calculated by dividing the gravitational force acting on the cable car by its mass. The result is then multiplied by the length of the hill and the coefficient of friction between the cable car and the track.

## 2. What is the significance of the 4122N braking force in the calculation?

The 4122N braking force is the force required to stop the cable car from moving down the hill. This force is an important component in the calculation of the cable car's speed on a hill descent as it determines the rate at which the cable car will decelerate.

## 3. How does the length of the hill affect the cable car's speed on a hill descent?

The longer the hill, the greater the distance the cable car has to travel, resulting in a higher speed. This is because the cable car has more time to accelerate due to the force of gravity acting on it.

## 4. What role does the coefficient of friction play in the calculation of the cable car's speed?

The coefficient of friction is a measure of the resistance between the cable car and the track. A higher coefficient of friction means that there is more resistance, resulting in a slower speed. Conversely, a lower coefficient of friction means less resistance and a higher speed for the cable car.

## 5. Are there any other factors that can affect the cable car's speed on a hill descent?

Aside from the factors mentioned above, other factors that can affect the cable car's speed on a hill descent include the weight of the cable car, any additional forces acting on the cable car, and the shape and angle of the hill.

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