Particle subject to position dependent force

AI Thread Summary
A particle with mass m and total energy E is influenced by a position-dependent force F(x) = ξx^4. To find the velocity v as a function of position x, the work-energy theorem is applied, stating that the net work done equals the change in kinetic energy. The work done by the force as the particle moves from x=0 to x(t) is calculated using the integral of F(x). This integral is set equal to the change in kinetic energy, allowing for the expression of v(x). The final step involves sketching the function v(x) to illustrate the particle's motion.
jimz
Messages
12
Reaction score
0

Homework Statement


A particle with total energy E and mass m is subject to a force F(x)=\xi x^4. Find the velocity v of the particle as a function of the position x, and sketch a phase diagram for the motion.

Homework Equations


T=\frac{1}{2}m\dot{x}^2

U=constant

F=m\ddot{x}

The Attempt at a Solution


x=\sqrt[4]{\xi m \ddot{x}}

Not sure where to go from here, or what the phase diagram axes should be. Do I just take the time derivative of x and that's my velocity?
 
Physics news on Phys.org
Use the work-energy theorem.
 
ehild said:
Use the work-energy theorem.
"The net work done by all the forces acting on a body equals the change in its kinetic energy."
Not seeing it...
 
Assume one-dimensional motion along x. You need the velocity of the particle as function of the position: v(x). At t=0 let x=0 and the kinetic energy=E. During some time period t, the displacement of the particle is x(t) and the change of KE is:

\Delta E = 1/2 mv(x)^2-E

The particle is subjected to a force of form

F(x) = \xi x^4.

The work done by this force while the particle moves from position x=0 to some x(t) is

W=\int_0^{x(t)}{F(x)dx}=\int_0^{x(t)}{\xi x^4dx}

Calculate the integral, make it equal to the change of KE, express v(x), sketch v(x) as function of x.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top