AdverseDeviant Posted February 12, 2011 Share Posted February 12, 2011 A particle of mass m is subjected to a force acting in the x-direction, Fx = (3.47 + 0.405x) N. Find the work done by the force as the particle moves from x = 0 to x = 4.79 m? Link to comment Share on other sites More sharing options...
Yoofie Posted February 12, 2011 Share Posted February 12, 2011 Is Fx a function? Or is this a simple algebraic equation? Without some sort of context, my guess would be to simply substitue the numbers into the equation. Link to comment Share on other sites More sharing options...
AdverseDeviant Posted February 12, 2011 Author Share Posted February 12, 2011 thats all it gave me. i tried substituting the difference in x then doing F*deltaX but it was wrong.. Link to comment Share on other sites More sharing options...
Samurizer Posted February 12, 2011 Share Posted February 12, 2011 Fx is a function with respect to x. That means that the force F will change as the value of x changes. Assuming that the particle moves in a straight line, the solution is: W = ∫ F ? dx Integrate from x = 0 to x = 4.79. Link to comment Share on other sites More sharing options...
-Himanshu- Posted February 12, 2011 Share Posted February 12, 2011 Is the answer 21.27 J? Link to comment Share on other sites More sharing options...
ArKeYa Posted February 12, 2011 Share Posted February 12, 2011 Fx is a function with respect to x. That means that the force F will change as the value of x changes. Assuming that the particle moves in a straight line, the solution is: W = ∫ F • dx Integrate from x = 0 to x = 4.79. This is the correct solution/equation. The TS should be able to integrate the simple linear equation by himself. It doesn't matter how the particle moved because it is only the work of the Force F we're after, which only applies in the x-direction. Link to comment Share on other sites More sharing options...
AdverseDeviant Posted February 12, 2011 Author Share Posted February 12, 2011 yes it is thanks. one more if yall dont mid A constant force, $\vec{F}$ = (2.57, -2.41, 1.13) N, acts on an object of mass 11.9 kg, causing a displacement of that object by $\vec{r}$ = (3.93, 3.91, -3.61) m. What is the total work done by this force? i tried sqrt(Fx^2 + Fy^2 + Fz^2) = Fnet then sqrt(Rx^2 + Ry^2 + Rz^2) = Rnet then Rnet * Fnet = Work but that didn't work.. then i tried Rnet * Fnet * Mass but taht didnt work either Link to comment Share on other sites More sharing options...
Samurizer Posted February 12, 2011 Share Posted February 12, 2011 yes it is thanks. one more if yall dont mid A constant force, $\vec{F}$ = (2.57, -2.41, 1.13) N, acts on an object of mass 11.9 kg, causing a displacement of that object by $\vec{r}$ = (3.93, 3.91, -3.61) m. What is the total work done by this force? i tried sqrt(Fx^2 + Fy^2 + Fz^2) = Fnet then sqrt(Rx^2 + Ry^2 + Rz^2) = Rnet then Rnet * Fnet = Work but that didn't work.. then i tried Rnet * Fnet * Mass but taht didnt work either You need the component of the force that is PARALLEL to the direction that the mass moves in. Use trigonometry to resolve the component of the force in the direction in which the mass is moving, and multiply that component by the distance. Link to comment Share on other sites More sharing options...
-Himanshu- Posted February 12, 2011 Share Posted February 12, 2011 Is the answer -3.4 J? Link to comment Share on other sites More sharing options...
AdverseDeviant Posted February 12, 2011 Author Share Posted February 12, 2011 yes it is. how did you get that? Link to comment Share on other sites More sharing options...
ArKeYa Posted February 12, 2011 Share Posted February 12, 2011 W = F (dot) d. Or W = dot product of constant force and displacement. Look up on how to do dot products on vectors. Link to comment Share on other sites More sharing options...
-Himanshu- Posted February 12, 2011 Share Posted February 12, 2011 Work done is F.dS. It is the dot product of force vector and displacement vector. You just need to calculate the dot product of the two vectors. Since it is already given in the x,y,z component form, you just have to multiply the x,y,z components of both the vectors and add them. Link to comment Share on other sites More sharing options...
Samurizer Posted February 12, 2011 Share Posted February 12, 2011 W = F (dot) d. Or W = dot product of constant force and displacement. Look up on how to do dot products on vectors. Apparently I'm terrible at maths. This is the better solution. :D Link to comment Share on other sites More sharing options...
AdverseDeviant Posted February 12, 2011 Author Share Posted February 12, 2011 okay i got it. thanks again Link to comment Share on other sites More sharing options...
+Warwagon MVC Posted February 12, 2011 MVC Share Posted February 12, 2011 For the record, my brains are currents oosing out the side of my head from this post. :yes: Link to comment Share on other sites More sharing options...
-Himanshu- Posted February 12, 2011 Share Posted February 12, 2011 For the record, my brains are currents oosing out the side of my head from this post. :yes: Why? It's simple physics. I am currently studying all this in 11th standard. Link to comment Share on other sites More sharing options...
+Warwagon MVC Posted February 12, 2011 MVC Share Posted February 12, 2011 Why? It's simple physics. I am currently studying all this in 11th standard. I bet it's easy if you new any kind of physics. Math hates me. :yes: Don't get me wrong, I can add, subtract, multiple and divide. Maybe this is why i'm not any good at programming :laugh: Link to comment Share on other sites More sharing options...
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