A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is f
ixed in place a distance dd above the first plate. Assume that dd is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.What is the tension in the cable? Neglect the weight of the plate.Express your answer in terms of the variables d, r, V, and constants ϵ0, pi.F=
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Answer:
μ = 0.125
Explanation:
To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.
Let's start working on the ramp
starting point. Highest point of the ramp
Em₀ = U = m h y
final point. Lower part of the ramp, before entering the rough surface
= K = ½ m v²
as they indicate that there is no friction on the ramp
Em₀ = Em_{f}
m g y = ½ m v²
v =
we calculate
v = √(2 9.8 0.25)
v = 2.21 m / s
in the rough part we use the relationship between work and kinetic energy
W = ΔK = K_{f} -K₀
as it stops the final kinetic energy is zero
W = -K₀
The work is done by the friction force, which opposes the movement
W = - fr x
friction force has the expression
fr = μ N
let's write Newton's second law for the vertical axis
N-W = 0
N = W = m g
we substitute
-μ m g x = - ½ m v²
μ =
Let's calculate
μ =
μ = 0.125