What's the speed after moving down the inclined plane?

Classical Mechanics Level 2

A body starts moving downwards (from rest) from the upper point of an inclined plane ( A A ). What is the formula of the velocity when the object arrives at the base of the plane ( B B )?

Details:

The velocity should be expressed in terms of:

  • The length of the inclined plane, L L .

  • Its angle, α \alpha .

  • The coefficient of friction on the plane, μ \mu .

  • The gravitational acceleration, g g .

v B = 2 l g ( sin ⁡ α − μ cos ⁡ α ) v_B=\sqrt{2lg(\sin\alpha-\mu\cos\alpha)} v B = 2 g h v_B=\sqrt{2gh} v B = l g ( sin ⁡ α − cos ⁡ α ) v_B=\sqrt{lg(\sin\alpha-\cos\alpha)} v B = ( l g ( sin ⁡ α − cos ⁡ α ) ) 2 v_B=(lg(\sin\alpha-\cos\alpha))^2

Victor Dumbrava by Victor Dumbrava , 17, Romania

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1 solution

Victor Dumbrava
Sep 24, 2017

At the very top of the plane, the object has no initial velocity, hence its energy is equal to its potential energy:

E A = P E A = m g l sin ⁡ α E_A=PE_A=mgl\sin\alpha

The friction force acts on the body as well, and the formula is:

F f = μ m g cos ⁡ α F_f=\mu mg\cos\alpha

The work of the friction force is:

W F f = − L ⋅ F f = − L μ m g cos ⁡ α W_{F_f}=-L\cdot F_f=-L\mu mg\cos\alpha

Applying the law of mechanical energy conservation:

E B = E A + W F f E_B=E_A+W_{F_f}

E B = m g L sin ⁡ α − L μ m g cos ⁡ α E_B=mgL\sin\alpha-L\mu mg\cos\alpha

Also note that at point B B , the object has no potential energy, hence:

E B = K E B = m v B 2 2 E_B=KE_B=\frac{mv_{B}^2}{2}

m v B 2 2 = m g L ( sin ⁡ α − μ cos ⁡ α ) \frac{mv_{B}^2}{2}=mgL(\sin\alpha-\mu \cos\alpha)

v B 2 2 = g L ( sin ⁡ α − μ cos ⁡ α ) \frac{v_{B}^2}{2}=gL(\sin\alpha-\mu \cos\alpha)

⟹ v B 2 = 2 g L ( sin ⁡ α − μ cos ⁡ α ) \implies v_{B}^2=2gL(\sin\alpha-\mu \cos\alpha)

v B = 2 L g ( sin ⁡ α − μ cos ⁡ α ) \boxed{v_{B}=\sqrt{2Lg(\sin\alpha-\mu \cos\alpha)}}

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