Question

Find analytically the velocity of the object at the end point of the inclined plane for a certain angle Ө

Answer 1

I don't know if there is other given information that's missing here, so I'll try to fill in the gaps as best I can.

Let m be the mass of the object and v₀ its initial velocity at some distance x up the plane. Then the velocity v of the object at the bottom of the plane can be determined via the equation

v² - v₀² = 2 a x

where a is the acceleration.

At any point during its motion down the plane, the net force acting on the object points in the same direction. If friction is negligible, the only forces acting on the object are due to its weight (magnitude w) and the normal force (mag. n); if there is friction, let f denote its magnitude and let µ denote the coefficient of kinetic friction.

Recall Newton's second law,

∑ F = m a

where the symbols in boldface are vectors.

Split up the forces into their horizontal and vertical components. Then by Newton's second law,

• net horizontal force:

∑ F = n cos(θ + 90º) = m a cos(θ + 180º)

→  - n sin(θ) = - m a cos(θ)

→  n sin(θ) = m a cos(θ) ……… [1]

• net vertical force:

∑ F = n sin(θ + 90º) - w = m a sin(θ + 180º)

→   n cos(θ) - m g = - m a sin(θ)

→   n cos(θ) = m (g - a sin(θ)) ……… [2]

where in both equations, a is the magnitude of acceleration, g = 9.80 m/s², and friction is ignored.

Then by multiplying [1] by cos(θ) and [2] by sin(θ), we have

n sin(θ) cos(θ) = m a cos²(θ)

n cos(θ) sin(θ) = m (g sin(θ) - a sin²(θ))

m a cos²(θ) = m (g sin(θ) - a sin²(θ))

a cos²(θ) + a sin²(θ) = g sin(θ)

a = g sin(θ)

and so the object attains a velocity of

v = √(v₀² + 2 g x sin(θ))

If there is friction to consider, then f = µ n, and Newton's second law instead gives

• net horizontal force:

∑ F = n cos(θ + 90º) + f cos(θ) = m a cos(θ + 180º)

→   - n sin(θ) + µ n cos(θ) = - m a cos(θ)

→   n sin(θ) - µ n cos(θ) = m a cos(θ) ……… [3]

• net vertical force:

∑ F = n sin(θ + 90º) + f sin(θ) - w = m a sin(θ + 180º)

→   n cos(θ) + µ n sin(θ) - m g = - m a sin(θ)

→   n cos(θ) + µ n sin(θ) = m g - m a sin(θ) ……… [4]

Then multiply [3] by cos(θ) and [4] by sin(θ) to get

- n sin(θ) cos(θ) + µ n cos²(θ) = - m a cos²(θ)

n cos(θ) sin(θ) + µ n sin²(θ) = m g sin(θ) - m a sin²(θ)

and adding these together gives

µ n (cos²(θ) + sin²(θ)) = m g sin(θ) - m a (cos²(θ) + sin²(θ))

µ n = m g sin(θ) - m a

m a = m g sin(θ) - µ n

m a = m g sin(θ) - µ m g cos (θ)

a = g (sin(θ) - µ cos (θ))

and so the object would instead attain a velocity of

v = √(v₀² + 2 g x (sin(θ) - µ cos (θ)))