P16.9 5.0 ?? is connected to a spring for which k-780 N/ m. The block A 74.0-g b
ID: 306245 • Letter: P
Question
P16.9 5.0 ?? is connected to a spring for which k-780 N/ m. The block A 74.0-g block carrying a charge Q 3 lies on a frictionless, horizontal surface and is immersed in a uniform electric field of magnitude E -4.86 x 104 N/ C directed as shown in Figure P16.9. If the block is released from rest when the spring is unstretched (x - O), (a) by what maximum distance does the block move from its initial position? (b) Find the subsequent equilibrium position of the block and the amplitude of its motion. (c) Using conservation of energy, find a symbolic relationship giving the potential difference between its initia position and the point of maximum extension in terms of the spring constant k, the amplitude A, and the charge Q.Explanation / Answer
(a) Electrical energy will be converted into the potential energy of the spring
F*x = (1/2)kx2
where x is the displacement
F is electric force = qE = 35*10-6*4.86*104 = 1.7 N
Now putting all the values in above equation
1.7 = (1/2)*78*x
x = 0.0436 m = 4.36 cm
(b) We know that at the equm position
F = kX
1.7 = 78 *X
X = 0.0218 m = 2.18 cm
(c)
From conservation of energy we know that
(1/2)kx2 = Fx
(1/2)kx =qE
x = 2qE/k
We know that V = E/x
x = V/E
putting the value in above equation
V/E = 2qE/k
V = 2qE2/k
where V is potential difference
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