cclement524
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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.