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.
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)
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)
To solve this problem, we can use the principles of classical mechanics, specifically the laws of motion and the concept of inclined planes.
i) The first step is to calculate the acceleration of the particle as it moves down the inclined plane. The net force acting on the particle is the component of the gravitational force parallel to the plane minus the force of friction. The equation for the net force is given by:
Net force = m * acceleration
where m is the mass of the particle.
The component of the gravitational force parallel to the plane is m * g * sin(θ), where θ is the angle of inclination of the plane. In this case, θ = tan^(-1)(4/3). Therefore, the gravitational force component is m * g * sin(tan^(-1)(4/3)).
The frictional force is given by the equation: frictional force = coefficient of friction * Normal force, where the Normal force is given by Normal force = m * g * cos(θ), where θ is the angle of inclination of the plane. In this case, θ = tan^(-1)(4/3). Therefore, the Normal force is m * g * cos(tan^(-1)(4/3)).
Substituting all the given values, we can write the equation for the net force as:
m * acceleration = m * g * sin(tan^(-1)(4/3)) - (1/3) * m * g * cos(tan^(-1)(4/3))
Simplifying the equation and canceling out the mass of the particle, we get:
acceleration = g * [sin(tan^(-1)(4/3)) - (1/3) * cos(tan^(-1)(4/3))]
Now, we can use the equation of motion: v^2 = u^2 + 2 * a * s, where v is the final velocity, u is the initial velocity (which is zero in this case since the particle is released from rest), a is the acceleration, and s is the distance traveled.
Since the particle is released from rest, u = 0, and the equation simplifies to:
v^2 = 2 * a * s
Now we can substitute the known values:
v^2 = 2 * [g * (sin(tan^(-1)(4/3)) - (1/3) * cos(tan^(-1)(4/3))))] * s
After substituting the values, calculate the square root of both sides of the equation to find the speed of the particle.
ii) To find the time taken to move from P to Q, we can use the equation of motion: v = u + a * t, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time.
Since the particle is released from rest, u = 0, and the equation simplifies to:
v = a * t
Now we can substitute the known values:
v = [g * (sin(tan^(-1)(4/3)) - (1/3) * cos(tan^(-1)(4/3))))] * t
Rearranging the equation to solve for t:
t = v / [g * (sin(tan^(-1)(4/3)) - (1/3) * cos(tan^(-1)(4/3))))]
After substituting the values, you can calculate the time taken to move from P to Q.