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

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The discussion revolves around solving physics problems related to work and roller coaster dynamics. A user seeks help with a problem involving a crate on an incline, where the work done by gravity is calculated using the formula Work = Fs Cos(x). The conversation highlights the importance of understanding angles and the concept of negative work when the direction of force opposes motion. Additional problems discussed include calculating work done by a varying force and the energy transformations in a spring system. The user expresses frustration over grasping these concepts, indicating a struggle with the material despite receiving guidance.
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I am a little confused by this problem. If someone could explain to me how to solve it, it would be greatly appreciated.

A 100 kg crate is on a rough surface inclined at 30 degrees. A constant external force P= 800N is applied horizontally to the crate. The force pushes the crate a distance of 3.0m up the incline, in a time interval of 5.2s, and the velocity changes from V1= 1.4 m/s to V2= 2.8 m/s. The work done by the weight is closest to:
A)+1400J
B)zero
C)-400J
D)-1400J
E)+400J
 
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Welcome to PF Williams,

Could you please any working or thoughts you have on the question.

Regards,
~Hoot
 
I just don't know where to start.
 
What is the equation for work done?

HINT: Alot of the information given in the question is superfluous.

Regards,
~Hoot
 
Work done = Fs Cos(x)

I kind of thought that there was some information in there that is irrelevant to solving the problem.
 
williams31 said:
Work done = Fs Cos(x)

That's the one, so in this case the force is weight, which is mg. And you have the distance traveled and you can calculate your angle (take care), therefore you can obtain the work done by gravity.

Regards,
~Hoot
 
Hootenanny said:
That's the one, so in this case the force is weight, which is mg. And you have the distance traveled and you can calculate your angle (take care), therefore you can obtain the work done by gravity.

Regards,
~Hoot
So that would be:
100(9.8)(3.0)cos(30)?

Cause that does not equal any of the choices.
 
williams31 said:
So that would be:
100(9.8)(3.0)cos(30)?

Cause that does not equal any of the choices.

You are almost right. Think about your angle, a sketch may be useful. I get an answer of -1471.5 J

Regards,
~Hoot
 
Hootenanny said:
You are almost right. Think about your angle, a sketch may be useful. I get an answer of -1471.5 J

Regards,
~Hoot
Well thanks for the help but I still do not understand this problem. I am pretty much screwed anyway. I have a take home test due tonight that I haven't been able to start until now. I have to somehow work out 10 problems when I can't even do this simple one. But thanks for helping.
 
  • #10
williams31 said:
Well thanks for the help but I still do not understand this problem. I am pretty much screwed anyway. I have a take home test due tonight that I haven't been able to start until now. I have to somehow work out 10 problems when I can't even do this simple one. But thanks for helping.

If you use trigonometry you will find that it becomes cos60

Regards,
~Hoot
 
  • #11
Hootenanny said:
If you use trigonometry you will find that it becomes cos60

Regards,
~Hoot
Thank you. But I still don't understand how that ends up being negative.
 
  • #12
Work done is defined as force multiplied by the distance moved in the direction of that force. The weight acts down, but the crate is moving upwards. Therefore, the work done is negative.

Regards,
~Hoot
 
  • #13
Wow, I really am an idiot. Can't believe I missed that. I have a question about another problem too.

An object of mass 2 kg is repelled from the origin by a force in the +x-direction whose magnitude varies with x according to F=(7Nxm^2)x^-2. How much work is done by this force when the object moves from x=2 to x=3? (Be sure to say whether this work is positive or negative.)

Do I just substitute for x for both and then take the difference?
 
  • #14
williams31 said:
Wow, I really am an idiot. Can't believe I missed that.

Everyone's done it atleast once before :biggrin:

williams31 said:
Do I just substitute for x for both and then take the difference?

No, you can't do that because the force is not constant. You have to integrate your function between the limits i.e.

wd = \int^{3}_{2} \frac{7}{x^2} \;\;dx

Regards,
~Hoot
 
  • #15
Hootenanny said:
Everyone's done it atleast once before :biggrin:



No, you can't do that because the force is not constant. You have to integrate your function between the limits i.e.

wd = \int^{3}_{2} \frac{7}{x^2} \;\;dx

Regards,
~Hoot
Well thank you again but I think I am giving up on this test. No matter how hard I try I cannot grasp the concept of this stuff. My head is killing me right now and I've only completed 5 out of 12 problems.
 
  • #16
Well, if you need anymore help don't hesitate to come back.

Regards,
~Hoot
 
  • #17
Hootenanny said:
Well, if you need anymore help don't hesitate to come back.

Regards,
~Hoot
Well I do need a lot of help but I feel like I would be a bother by asking for help on so many problems. Especially when it takes me forever to just comprehend one of them.
 
  • #18
williams31 said:
Well I do need a lot of help but I feel like I would be a bother by asking for help on so many problems. Especially when it takes me forever to just comprehend one of them.

Not at all, all the homework helpers (and others) here would take pleasure in helping you. As long as you are willing to put the work in we will guide you through the questions.

Regards,
~Hoot
 
  • #19
Hootenanny said:
Everyone's done it atleast once before :biggrin:



No, you can't do that because the force is not constant. You have to integrate your function between the limits i.e.

wd = \int^{3}_{2} \frac{7}{x^2} \;\;dx

Regards,
~Hoot
Im still not really sure what this question is asking...
 
  • #21
Hootenanny said:
Perhaps this page may shed some light on the matter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/wint.html#wi
Well I think I will just leave that one blank. I have no clue what is going on.

This is another one of the questions:
A 5.00 kg block is moving at 5.00 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall. After the block collides with the spring, the spring is compressed a maximum distance of .68 m. What is the speed of the block when the spring is compressed to one-half of the maximum distance?
 
  • #22
What do you know abou the kinetic energy when the spring is compressed to half is amplitude?

~Hoot
 
  • #23
Hootenanny said:
What do you know abou the kinetic energy when the spring is compressed to half is amplitude?

~Hoot
It is twice as much?
 
  • #24
I meant in relation to the intial kinetic energy. The total amount of energy must remain constant;

E_{t} = E_{k} + E_{p}

HINT: As the spring compresses it gains potential energy (E_{p}). At it's maximum compression E_{p} = E_{t} \Rightarrow E_{k} = 0.
 
  • #25
Hootenanny said:
I meant in relation to the intial kinetic energy. The total amount of energy must remain constant;

E_{t} = E_{k} + E_{p}

HINT: As the spring compresses it gains potential energy (E_{p}). At it's maximum compression E_{p} = E_{t} \Rightarrow E_{k} = 0.
I am lost.
 
  • #26
When the spring is being compressed kinetic energy is being converted into (elastic) potential energy. When the spring is fully compressed all the kinetic energy will be converted into potential. 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.

Do you follow?

~Hoot
 
  • #27
Hootenanny said:
When the spring is being compressed kinetic energy is being converted into (elastic) potential energy. When the spring is fully compressed all the kinetic energy will be converted into potential. 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.

Do you follow?

~Hoot
Ok now I understand.
 
  • #28
I just don't get how you can find the speed.
 
  • #29
williams31 said:
I just don't get how you can find the speed.

What is the equation for kinetic energy?
 
  • #30
1/2mv^2...
 
  • #31
Yes, so you have the intial velocity, this allows you calculate the intial kinetic energy. Can you go from here?

~Hoot
 
  • #32
Hootenanny said:
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??
 
  • #33
Yep that's correct. The next step;

Hootenanny said:
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
 
  • #34
Hootenanny said:
Yep that's correct. The next step;



~Hoot
So now the kinetic engergy = 31.25...
 
  • #35
williams31 said:
So now the kinetic engergy = 31.25...

You've got it
 
  • #36
So now its:
31.25=1/2(5)v^2 and I just have to solve for v?
 
  • #37
Yeah, that's right
 
  • #38
Hootenanny said:
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
 
  • #39
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
 
  • #40
I came up with .96 which I guess would be answer A.
 
  • #41
williams31 said:
I came up with .96 which I guess would be answer A.

That's correct.

Regards,
~Hoot
 
  • #42
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.
 

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