Work/Kinetic Theory Problem

[williams31]

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

 

[Hootenanny]

Welcome to PF Williams,

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

Regards,
~Hoot

 

[williams31]

I just don't know where to start.

 

[Hootenanny]

What is the equation for work done?

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

Regards,
~Hoot

 

[williams31]

Work done = Fs Cos(x)

I kind of thought that there was some information in there that is irrelevant to solving the problem.

 

[Hootenanny]

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

 

[williams31]

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.

 

[Hootenanny]

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

 

[williams31]

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.

 

[Hootenanny]

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

 

[williams31]

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.

 

[Hootenanny]

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

 

[williams31]

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?

 

[Hootenanny]

Wow, I really am an idiot. Can't believe I missed that.

Everyone's done it atleast once before

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

 

[williams31]

Everyone's done it atleast once before



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.

 

[Hootenanny]

Well, if you need anymore help don't hesitate to come back.

Regards,
~Hoot

 

[williams31]

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.

 

[Hootenanny]

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

 

[williams31]

Everyone's done it atleast once before



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...

 

[Hootenanny]

Perhaps this page may shed some light on the matter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/wint.html#wi

 

[williams31]

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?

 

[Hootenanny]

What do you know abou the kinetic energy when the spring is compressed to half is amplitude?

~Hoot

 

[williams31]

What do you know abou the kinetic energy when the spring is compressed to half is amplitude?

~Hoot

It is twice as much?

 

[Hootenanny]

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 $E_{p} = E_{t} \Rightarrow E_{k} = 0$.

 

[williams31]

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.

 

[Hootenanny]

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

 

[williams31]

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.

 

[williams31]

I just don't get how you can find the speed.

 

[Hootenanny]

I just don't get how you can find the speed.

What is the equation for kinetic energy?

 

[williams31]

1/2mv^2...

 

[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