Pls solve this question.

Asked by suchananag2002 | 22nd Mar, 2020, 10:10: AM

Expert Answer:

Forces acting on a charged ball that is slipping down on an inclined plane is shown in figure.
 
mg is weight, gravitational force that is resolved into two components.
One component mgsin30 which is parallel to inclined plane driving the ball down.
 
Other component mgcos30 acting perpendicular to surface of inclined plane.
 
qE is the electrostatic force, where q is charge of the ball and E is electric field.
This electrostatic force also is resolved into two components.
 
One component, qEcos30 is parallel to inclined plane surface that oppose the motion of the ball.
 
Other component qEsin30 which is perpendicular to the inclined plane surface.
 
Hence the normal reaction force = N = mgcos30 + qEsin30
 
Friction force μN = μ ( mgcos30 + qEsin30 )
 
Hence the net force driving the ball down = { mg sin30 - qE cos30 - μ ( mgcos30 + qEsin30 ) } ..............(1)
 
electro static force = qE = 0.01 × 100 = 1 N 
 
Acceleration in the direction of ball's slipping motion = force / mass
 
using qE= 1 N and given value of friction coefficient μ = 0.2 ,
 
we get acceleration a from eqn.(1) as ,  a = 2.237 m/s2
 
 
Time taken t by the ball to slip down to bottom is obtained from " S = (1/2) a t2 "
 
where S is length of incline plane surface, i.e. 2 m 
( height = 1m , angle 30o , hence length of inclined plane = 2 m )
 
begin mathsize 14px style t space equals space square root of fraction numerator 2 space S over denominator a end fraction end root space equals space square root of fraction numerator 2 cross times 2 over denominator 2.237 end fraction end root end style = 1.337 s

Answered by Thiyagarajan K | 22nd Mar, 2020, 05:40: PM

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