A 79.0-g block carrying a charge Q = 34.0 mu C is connected to a spring for whic

Question

A 79.0-g block carrying a charge Q = 34.0 mu C is connected to a spring for which k = 75.0 N/m. The block lies on a frictionless, horizontal surface and is immersed in a uniform electric field of magnitude E = 4.82 Times 10^4 N/C directed as shown in the figure below. The block is released from rest when the spring is unstretched (x = 0). By What maximum distance does the block move from its initial position? . Find the subsequent equilibrium position of the block and the amplitude of its motion. (Indicate the direction with the sign of your answer.) position

Explanation / Answer

(a)   Using energy balance equation,

Change in potential energy due to charge = Gain in potential energy in spring

  Q*E*x  = (1/2)*k*x^2

  =>   Q*E = (1/2)*k*x

   => (34*10^-6)*(4.82*10^4) = (1/2)*75*x => x = 0.0437 m = 4.37 cm

(b) The equilibrium position of the block => x = 0

The amplitude of the periodic motion:   A = 4.37 cm

  

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