Question
A particle is released from rest at a point P on an inclined plane. The inclination of the plane is tan"1 (4/3) to the horizontal. If the coefficient of friction between the particle and the plane is 1/3, find i) the speed of the particle when it passes Q, where the distance of PQ is 10 m.
ii) the time to move from P to Q.
ii) the time to move from P to Q.
Answers
Let X = 10 m be the distance from P to Q. There will be energy loss due to friction as the particle slides from P to Q.
The angle from horizontal is A = arctan(4/3)
= 53.13 deg
sin A = 4/5 cosA = 3/5
i)
The friction force opposing motion is
Ff = M g cosA*(1/3) = (1/5)*M*g
Kinetic energy at Q = (Potential Energy Loss) - (Frictional work)
(1/2)MV^2 = M g L sinA - Ff*L
= M*g*L (3/5) - (1/5)M*g*L
= (2/5)*M*g*L
The M's cancel. Solve for V when L = 10
ii) Divide L by the average velocity from i)
The angle from horizontal is A = arctan(4/3)
= 53.13 deg
sin A = 4/5 cosA = 3/5
i)
The friction force opposing motion is
Ff = M g cosA*(1/3) = (1/5)*M*g
Kinetic energy at Q = (Potential Energy Loss) - (Frictional work)
(1/2)MV^2 = M g L sinA - Ff*L
= M*g*L (3/5) - (1/5)M*g*L
= (2/5)*M*g*L
The M's cancel. Solve for V when L = 10
ii) Divide L by the average velocity from i)
there's a mistake in substitution:
The friction force opposing motion is
Ff = M g cosA*(1/3) = (1/5)*M*g
Kinetic energy at Q = (Potential Energy Loss) - (Frictional work)
(1/2)MV^2 = M g L sinA - Ff*L
= M*g*L (3/5) - (1/5)M*g*L
= (2/5)*M*g*L
its supposed to be M*g*L (4/5) when you substitute sin A, not M*g*L (3/5)
The friction force opposing motion is
Ff = M g cosA*(1/3) = (1/5)*M*g
Kinetic energy at Q = (Potential Energy Loss) - (Frictional work)
(1/2)MV^2 = M g L sinA - Ff*L
= M*g*L (3/5) - (1/5)M*g*L
= (2/5)*M*g*L
its supposed to be M*g*L (4/5) when you substitute sin A, not M*g*L (3/5)