A 0.3 kg ball is released in an inclined plane as shown in the figure below. In order for it to stop at point B, what is the magnitude of the kinetic friction force?
A 0.3 kg ball is released in an inclined plane as shown in the figure below. In order for it to stop at point B, what is the magnitude of the kinetic friction force?
Homework Answers
Given, mass of ball, m = 0.3 kg
From given figure, sin(35ο) = AD/AC, => AC = 8.72m
sin(25ο) = BE/BC, => BC = 4.73m
total distance, d = 8.72 + 4.73 = 13.45m
lets assume, kinetic friction force = fx
the energy used for friction, E = fx x d
change in potential energy of ball, P = mg(AD-BE)
= 0.3x9.8x(5-2)
= 8.82 J
Here, E = P = 8.82 J
therefore, fx = E/d = 8.82/13.45 = 0.656 N
In order for it to stop at point B, 0.656 N kinetic friction force is needed.
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