CBSE Class 12-science Answered
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Asked by suchananag2002 | 22 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 )
= 1.337 s
Answered by Thiyagarajan K | 22 Mar, 2020, 05:40: PM
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