Energy question help?

[TMJ]

As a fully loaded roller-coaster car of total mass 450kg approaches point A with a speed of 12m/s, the power fails and it rolls freely down the track. Ignore friction.
a)calculate the kinetic energy    I got 36400J 
b) Determine the speed of the loaded car at point B and point C, need help?
Point A is at a height of 20m and point B is at a height of 0m and point C is at a height of 8m.

[Twanny666]

Kinetic Energy = .5 x M x (V)squared...
Thus it should be .5 x 450 x (12)squared which equals 32,400.

[Twanny666]

Does the cart go down to point b then rise back up to point C?

[TMJ]

Yeah thats correct down to B rises to C

[Twanny666]

The Total energy at point A is KE + PE = .5mvsquared + M  G  H = 32,400 + 88,200 = 120,600

The Energy at Point B is only Kinetic Energy as there is no height.

The law of Conservation of Energy states that the Sum of Energy at A Should equal the Sum of Energy at B.

Thus, The Kinetic Energy at must total 120,600.  Sub in M and solve for v and u get a velocity of approximately 23.15m/ssquared

To determine the speed at point C we must first determine the Kinetic energy at that point.

We know the Total Energy will once again be 120,600. But seeing as it is off the ground some of it will be Potential Energy (MGH), this amounts to 35,280J.

Thus the Kinetic Energy will equal 120,600-35,280 = 85,320 J. Now use the kinetic energy equation to find the velocity.. and you should get a velocity of approx. 19.47m/s squared.

Hope this helps.. I'm a little bit rusty as I did Kinematics at the start of the year... Haven't started revising for exam.

[Guest]

I found a different way of solving this question if you are comfortable with with the standard motion equations.
It's a slightly longer method, but fool-proof in the sense that it still works if you made an error with the kinetic energy calculation in part (a).

First find the velocity of Cart reaching point B from point A.
Use: v²=u²+2ax
v²=12²+(2*10*12)
v= sqrt(384)=19.60

Then find the velocity of Cart reaching point C from point B.
v²=sqrt(384)²+ 2*10*8
v²=sqrt(544)=23.32ms/s.

[SocialRhubarb]

Divide both sides by m, multiply both sides by 2, and substitute F=ma:





THEY'RE THE SAME FORMULA : OOOOO.

[BasicAcid]

Divide both sides by m, multiply both sides by 2, and substitute F=ma:





THEY'RE THE SAME FORMULA : OOOOO.

Hey SocialRhubarb, do you have a tutor that teaches you all of these connections or does your teacher at school teach you this stuff?
Or do you look it up yourself? If so, do you have any links/sources that I could read/use? I find the connections like this really interesting.
It's a pity they don't teach us neither at school or in the Heinemann textbook.

[Guest]

Hey SocialRhubarb, do you have a tutor that teaches you all of these connections or does your teacher at school teach you this stuff?
Or do you look it up yourself? If so, do you have any links/sources that I could read/use? I find the connections like this really interesting.
It's a pity they don't teach us neither at school or in the Heinemann textbook.

All he did was substitute and rearrange (it's maths knowledge)
I suppose what he should be commended for is seeing the link between both, instead of trying to find a formula which would allocate each and every variable perfectly. In other words, he turned theory into practicality in the use of equations.

[Guest]

Oh, and on that note, try using Microsoft Excel when you want to construct an equation. You will have to think independently and utilise equations you already know to construct one that works succinctly. By this I mean an equation which there is less manual rearranging, as you must ultimately only provide the computer with a single variable being equal to something else. Hard to explain really, and equally as hard to believe. I found this out for myself while doing the last EPI, where there were literally dozens of theoretical calculations that depended on formula. Constructing one formula that worked for all was the way to go.