[Work/Energy] a particle inside a field

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SUMMARY

The discussion focuses on the work done by a particle moving in a field with potential energy defined as U = -kx + c, where c is a constant. The particle experiences a damping force F = -b\vec{v} and has an initial velocity of v = V_0\hat{y}. The net force acting on the particle is derived as F = ke^{-\frac{b}{m}t}\hat{x} - bV_0e^{-\frac{b}{m}t}\hat{x}. The power of the damping force when acceleration is zero is calculated as P = \frac{k^2}{b}, confirming the relationship between work and power in this context.

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Homework Statement



a particle is moving in a field where it's potential energy is U= -kx+c , c is a constant

it is also experiencing a damping force F=-b\vec{v}
it has an initial velocity v=V_0\hat{y}

d. what is the work done by the net force throughout all of the particle's movment?
e. what is the power of the damping force when the acceleration of the particle is 0?


Homework Equations



in first the rest of the questions I have found :
the velocity:
v_x=-\frac{k}{b} (1-e^{-{frac{b}{m}t})
v_y= V_0 e^{-\frac{b}{m}t} \hat{y}

the velocity when a=0:
v_max = -\frac{k}{b}\hat{x}

the net force
F=ke^{-\frac{b}{m}t}\hat{x} -bV_0e^{-\frac{b}{m}t}\hat{x}

The Attempt at a Solution


d. on one hand the force acts "until" t=\infty so it should be infinity...
on the other hand the force will be very small so I'm positive it isn't infinity :/

e. the damping force at a=0 or t=\infty will be quite obviously
F=k\hat{x}
(that way it will be equal in magnitude and opposite in direction of the force causing the particle to move)

but we need the power so it's basically
P=Work per second=F\cdot \vec{v_max}
which is
P=\frac{k^2}{b}
am I wrong?
I used the velocity because it's how much x is changed per second - it ok by units, but is it right?
Thanks!
 
Last edited:
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you could use the work-energy theorem, or just integrate the net force on the path of the particle..
 

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