How Do You Solve These Physics Problems on Work and Roller Coaster Dynamics?

[Hootenanny]

Yes, so you have the intial velocity, this allows you calculate the intial kinetic energy. Can you go from here?

~Hoot

 

[williams31]

Yes, so you have the intial velocity, this allows you calculate the intial kinetic energy. Can you go from here?

~Hoot

K=1/2mv^2
K=1/2(5)(5)^2
K=62.5??

 

[Hootenanny]

Yep that's correct. The next step;

Therefore, when the spring is half compressed half of the kinetic energy will have been converted into potential energy, leaving the other half as kinetic.

~Hoot

 

[williams31]

Yep that's correct. The next step;



~Hoot

So now the kinetic engergy = 31.25...

 

[Hootenanny]

So now the kinetic engergy = 31.25...

You've got it

 

[williams31]

So now its:
31.25=1/2(5)v^2 and I just have to solve for v?

 

[Hootenanny]

Yeah, that's right

 

[williams31]

Yeah, that's right

Thanks...Now this next problem kind of confuses me. Because it seems like they have extra information but I am not really sure.

The force constant of a spring is 200N/m and its unstretched length is 16cm. The spring is placed inside a smooth tube that is 16 cm tall. A 0.72 kg disk is lowered onto the spring. An external force P pushes the disk down further, until the spring is 6.4 cm long. The external force is removed, the disk is projected upward and it emerges from the tube. The elastic potential energy of the spring is closest to:
A) .92J
B) .31J
C) .51J
D) .72J
E) .61J

 

[Hootenanny]

The force of a spring is given by

$$F = -kx$$

Therefore, the elastics potential energy stored in the spring is given by;

$$E_{p} = \int -kx \;\; dx = -\frac{1}{2}kx^2$$

Where x is the compression / extension. k is the spring constant.

Can you go from here? Again they have added additional information.

~Hoot

 

[williams31]

I came up with .96 which I guess would be answer A.

 

[Hootenanny]

I came up with .96 which I guess would be answer A.

That's correct.

Regards,
~Hoot

 

[williams31]

A roller coaster descends 35 meters in its intial drop and then rises 23 meters before going over the first hill. If a passenger at the top of the hill feels an apparent weight which is one half her normal weight, what is the radius of curvature of the first hill? Assume no frictional loss and neglect the speed of the roller coaster.

Now this one is really confusing to me. This is the first time I have encoutered a problem like this and I don't remember learning about something like this.