A 5kg block is carrying a charge Q=60 micro C is connected to a spring for which

Question

A 5kg block is carrying a charge Q=60 micro C is connected to a spring for which k=100N/m. The block lies on a frictionless horizontal track, and the system is immersed in a uniform electric field of magnitude E=500000 V/m.If the block is released from rest when spring is unstreached (at x=o), A) If the block is released from rest at x=0, how fast is it moving at x=0.25m? B) If instead of being released from rest, the block is shoved to the left so that it starts wih a velocity to the left of 2m/s, what maximum distance to the left of x=0 would the block reach? C) What maximum distance to the right of x=0 would the block reach? D) What would the new equilibirium location be? E) What if the block was shoved to the right giving it an initial velocity of 2m/s. What maximum distance to the right of x=0 would the block reach?

Explanation / Answer

A).

force by the electric field = qE = 60*10-6 *5*105 = 30 N (right direction)

from the Work Energy theorem,

work done= change in kinetic energy

30*0.25 - (1/2)*100*0.252=(1/2)*5*v2

solving this, we can get, velocity at x=0.25 m, v = 1.3229 m/s

B)

-30*x+ (1/2)*100*x2=(1/2)*5*22

or 50x2-30x-10=0

or 5x2-3x-1 =0

x1, x2 =0.8385 m, -0.2385 m

hence, maximum distance to the left of x=0 will be at x = -0.2385 m

C).

30*x- (1/2)*100*x2=(1/2)*5*1.32292

or 50x2-30x-4.375=0

or 5x2-3x-0.4375 =0

x1, x2 =0.7213 m, -0.1213 m

hence, maximum distance to the left of x=0 will be at x= 0.7213 m

D).

new equilibrium position will be at

kx=qE

100*x=30

x= 0.3 m

E).

30*x-(1/2)*100*x2=(1/2)*5*22

or 50x2-30x-10=0

or 5x2-3x-1 =0

x1, x2 =0.8385 m, -0.2385 m

maximum distance to the right of x=0 would the block reach= 0.8385 m

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