I agree it is a major assumption, but without actually seeing the lab we are unsure of the complete reaction between the two objects. I was just trying to use a broad approach to show how μ would be found if the pendulum reacted in that situation.
N=Normal
W= Weight
U= Work
We have the weight of the box (W) pushing down on the surface and the normal reaction for pushing up (N) . Will assume up is positive.
ΣFy=0
N-W=0
N=W
W=mg
If we assume the bob came to rest as it hit the box then we know the initial velocity of the box as 0.33 m/2 and the final velocity as 0 m/s (rest).
From this can we not take Kinetic(initial) + Work = Kinetic(final) T1 + U = T2. U=FΔx
½ mvi2 – FΔx = ½ mvf2
FΔx => FfΔx Ff...
I used 1/2io2L
1J=.5io2(3mH)
2J/(3mH)=io2
R=P/i2
R=1/(25.82)
R=.0015. (I'm sure this I wrong and we haven't covered in our notes how to proceed to my knowledge)
1. The problem statement, all variables, and given/known data
I have a single loop RL circuit with no source. I am given the the inductance of 3mH, as well as that energy storage is as follows 1 J @ t=0 and 100mJ @ t=1ms. No current is given. I have to find R. I don't need the answer just some...