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

[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?

Homework Equations

F = ma

The Attempt at a Solution

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

 

[Rake-MC]

Use conservation of energy.

$$U_g = mg\Delta h$$

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

 

[bluebear19]

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!

 

[Doc Al]

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.

 

[bluebear19]

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

 

[Doc Al]

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.

 

[bluebear19]

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!

 

[Doc Al]

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.

 

[bluebear19]

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??

 

[Doc Al]

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.

 

[bluebear19]

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?

 

[bluebear19]

nevermind I got it thank you sooooo much