A conducting rod slides down between two frictionless vertical copper tracks

[cclement524]

Homework Statement

A conducting rod slides down between two frictionless vertical copper tracks at a constant speed of 4.5 m/s perpendicular to a 0.56-T magnetic field. The resistance of the rod and tracks is negligible. The rod maintains electrical contact with the tracks at all times and has a length of 1.2 m. A 0.82-Ω resistor is attached between the tops of the tracks. Find the change in the gravitational potential energy that occurs in a time of 0.20 s.

Homework Equations

s = ut+ (1/2) gt^2

U = mgh

The Attempt at a Solution

I found the mass of the rod to be 0.2528777 kg

Here's my attempt at the answer, but I went wrong somewhere (I'm positive the mass I found is correct):

From the kinematic relations
s = ut+ (1/2) gt^2
= 0 +(1/2)gt^2
= (0.5)(9.80 m/s2)(0.20 s)^2
= 0.196 m
The change in the gravitational potential energy is
U = mgh
=( 0.2528777 kg)(9.80 m/s2)(0.196 m)
= 0.485727 J

My second attempt:

When it comes to finding ∆h I assumed free-fall (used 'g') ... but the rod was falling at constant speed
so I tried ∆h = 0.9 m
=(0.2528777 kg)(0.9 m)
= 0.20483

Neither of these are correct, but I am confused as to where I went wrong.

 

[supratim1]

there will be force due to electromagnetic induction. F = BIL

 

[cclement524]

Got it-- thank you!

 

[supratim1]

welcome.