As shown in the figure above, a negatively charged ball is placed at point A and
ID: 1545790 • Letter: A
Question
As shown in the figure above, a negatively charged ball is placed at point A and slides down the slope from rest. The area has a uniform electric field E = 60 N/C, pointing to the right. The mass of the ball is m = 15 kg and the charge is q = 40 C. When the ball reaches point B, it travels horizontally there after. The height from A to B is h = 40 m and the horizontal distance between A and B is d = 1 m. You can ignore friction and use g = 10 m/s2 for your calculation.
How much is the kinetic energy of the ball when it reaches at B (in the unit of J)? ______ J
What is the minimum strength of electric field E that can stop the particle right at Point B? ______ N/C
E 17 ATI hExplanation / Answer
From the given data
when the ball moves at rest from A to B on surface
the conservation of energy , gravitaional potential energy to kinetic energy
that is
mgh = 0.5*mv^2
v = sqrt(2gh)
V = sqrt(2*9.8*40) m/s
v = 28 m/s
the kinetic energy on the surface is k.e = 0.5*15*28^2 J = 5880 J
minimum strength of the electric field to stop the ball on surface is
the work done = change in k.e
work done by the electric force is W = F*d
E*q*d = 0.5*m(v2^2-v1^2)
E = 0.5*m(v2^2-v1^2)/(q*d)
E = 0.5*15(0-28^2)/(-40*1) N/C
E = 147 N/C
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