TranceAddict Forums (www.tranceaddict.com/forums)
- Chill Out Room
-- For you physics people..i need help with these problems

For you physics people..i need help with these problems

Question:

A 1.8 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F(x)=(2.4-x^2)i N, where x is in meters and the initial position of the block is x=0

(a) what is the kinetic energy of the block as it basses through x=2.0m?

Answer:

This is how i tried to solve it, since Vi(initial)=0 and Vf(final)= unknown, we can use the formula Vf^2=Vi+2a(X2-X1). But..Force = (2.4-x^2) and at x=2 it is (2.4-4)=-1.6....and f=ma so -1.6/1.8 will equal the acceleration which is -.89...but how the f is the acceleration negative when a horizontal force is applied to the box on a frictionless surface? i dont know where i went wrong with my calculatoins...

Also here is another question that i am having trouble with...

Question:

Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50m/s. At a certain location the conveyor belt moves for 2.0 m u an incline that makes an angle of 12degrees with the horizontal. Assume that the 2.0kg box rides on the belt without slipping

(a) At what rate is the force of the conveyor belt doing work on the box as the box moves up the 12degree incline.

Answer:

this is how i tried to solve it...

since m=2kg, a=0, theta= 12degree, and distance = 2m, and Vc(velocity at constant)=0.50...we have all the constants we need

So first Work=mgdcos(theta) = (2)(9.8)(2)(cos12)=38.343

and X2-X1 (change in distance)=(Vc)(time) sooo....2=38.343t and t = 4 seconds...

and Power = Work/change in time ...sooo 38.343/4 = 9.586 which is not the answer? help what am i doing wrong guys! thanks

hasnt there been a few threads recently with the same question?

Damn, TA isn't your afterschool homework class

These kind of threads may attract intelligent people to TA, and that's really the last thing we want!

im waiting for the thread saying, "OMG, I am taking the [insert SAT, GRE, or other standardized test] and I need the answer to this question...."


that would be sweet.

quote:

Originally posted by DJ-Fuq hasnt there been a few threads recently with the same question?

nope, same poster, different questions

you need to change your thinking:

first problem you are assuming that acceleration is constant when its not. if Force is given as a function of time, Acceleration is changing with time also.

second problem: think of the direction that the belt is moving relative to the direction that work is going. the belt is doing work against gravity. which way is the belt moving? which way is gravity pointing?

I haven't sold a single physics problem the last five years (i miss school... ), but I'll try
I hope you know this:
Kinetic energy (Final) - Kinetic energy (beginning) = Sum of all forces' works
Since you only have one force, and the box is not moving at the beginning, Kinetic Energy (final) = WF
When F is not constant, WF=the integral of F
which equals to [2,4*x-1/3*x^3]=G(x)
WF=G(2)-G(0)=2.1333
i'm not sure how to solve it without using integrals..
If i made a mistake, plz correct me

may be there should a Homework section on TA

Re: For you physics people..i need help with these problems

quote:

Originally posted by stk Question: A 1.8 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F(x)=(2.4-x^2)i N, where x is in meters and the initial position of the block is x=0 (a) what is the kinetic energy of the block as it basses through x=2.0m? Answer: This is how i tried to solve it, since Vi(initial)=0 and Vf(final)= unknown, we can use the formula Vf^2=Vi+2a(X2-X1). But..Force = (2.4-x^2) and at x=2 it is (2.4-4)=-1.6....and f=ma so -1.6/1.8 will equal the acceleration which is -.89...but how the f is the acceleration negative when a horizontal force is applied to the box on a frictionless surface? i dont know where i went wrong with my calculatoins...

You haven't done anything wrong with your calculations here because the force is not constant. So at the end it is negative but the object is still moving forward (although slowing down). You have to get the end velocity of the object, which you must do in this case by solving integral equations. When you do that, the energy equals mv^2/2. I'm too tired to do all the work for you now, but I might do it later. Really this is not nearly as easy problem as those ones you mentioned earlier.

quote:

Also here is another question that i am having trouble with... Question: Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50m/s. At a certain location the conveyor belt moves for 2.0 m u an incline that makes an angle of 12degrees with the horizontal. Assume that the 2.0kg box rides on the belt without slipping (a) At what rate is the force of the conveyor belt doing work on the box as the box moves up the 12degree incline. Answer: this is how i tried to solve it... since m=2kg, a=0, theta= 12degree, and distance = 2m, and Vc(velocity at constant)=0.50...we have all the constants we need So first Work=mgdcos(theta) = (2)(9.8)(2)(cos12)=38.343 and X2-X1 (change in distance)=(Vc)(time) sooo....2=38.343t and t = 4 seconds...

Think a bit. If the angle is 90 degrees then the force is at it's peak. And what's the cos of 90? 0. What's the sin of 90? 1. So the equation should be:

W=Fs=mgdsin(theta)=8.16

quote:

and Power = Work/change in time ...sooo 38.343/4 = 9.586 which is not the answer? help what am i doing wrong guys! thanks

Yes, P=W/t=2.04

Powered by: vBulletin
Copyright © 2000-2021, Jelsoft Enterprises Ltd.