Interesting Electric Field and Spring Problemf

AI Thread Summary
The discussion revolves around a physics problem involving a block with mass m and charge +Q connected to a spring in a uniform electric field. The maximum expansion of the spring is determined to be 2QE/k, while the equilibrium position of the block is found to be half of this maximum expansion, where the electric force equals the spring's restoring force. The block's motion is identified as simple harmonic, and the participants are working through the calculations to confirm the period of this motion. The relationship between the potential energy of the spring and the work done by the electric force is emphasized for solving part (a). This problem illustrates the interplay between electric forces and spring dynamics in a frictionless environment.
caesius
Messages
22
Reaction score
0

Homework Statement


A block having mass m and charge +Q is connected to an insulating spring having constant k. The block lies on a frictionless, insulating horizontal track, and the system is immersed in a uniform electric field of magnitude E as shown in the diagram. If the block is released from rest when the spring is unstretched (at x = 0),

(a) by what maximum amount does the spring expand?
(b) what is the equilibrium position of the block
(c) show that the block's motion is simple harmonic and determine its period
(d) repeat (a) assuming that the coefficient of kinetic friction between the block and the surface is mew K.

http://img513.imageshack.us/img513/3959/physprobzm4.png

Homework Equations


a few, F = qE = ma = -kx

The Attempt at a Solution



I'm only starting to attempt part (a), the block has a force QE to the right and a restoring force kx to the left.

So I equated the forces thus: QE = kx and solved for x to obtain x = QE/k.

This isn't the solution for (a), but it is for (b). The solution for (a) is 2QE/k

It's driving me crazy trying to find where the 2 came into calculations. Working backwards from (b) I see that the equilibrium position is half of the maximum expansion. But how so?
 
Last edited by a moderator:
Physics news on Phys.org
Right, that's the answer for B because at the equilibrium position, the restoring force of the spring is equal to the electric force. That's why at equilibrium, Fnet is 0 so the acceleration of the block is also zero.

For part A, consider the relationship between the potential energy of the spring at the point of furthest expansion and the work the electric force must do to get the block to that point.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top