A cube slides down a rough inclined plane of inclination θ with const

1.	A cube slides down a rough inclined plane of inclination θ with constant velocity. If this cube is projected up the plane with a velocity, v0 then at what distance along the inclined plane, the cube will come to roll?
Answer» Since the cube moves down the plane, then the force of friction equals the component of weight parallel to the inclined plane. i.e., f = mg sinθ When the block is moved up the inclined plane, then retarding force = mg sinθ + f = mg sinθ + mg sinθ = 2 mg sinθ Hence, a = {-2 mg sinθ}/{m} = -2 g sinθ Now, μ = v₀, v = 0; a = -2g sinθ, s = ? Using v² - u² = 2as ⇒ (0)² - (v₀)² = 2(-2 g sinθ) s or, s = {v₀²}/{4 g sinθ}

A cube slides down a rough inclined plane of inclination θ with constant velocity. If this cube is projected up the plane with a velocity, v0 then at what distance along the inclined plane, the cube will come to roll?

Answer»

Since the cube moves down the plane, then the force of friction equals the component of weight parallel to the inclined plane.

i.e., f = mg sinθ

When the block is moved up the inclined plane, then retarding force

= mg sinθ + f = mg sinθ + mg sinθ

= 2 mg sinθ

Hence, a = {-2 mg sinθ}/{m} = -2 g sinθ

Now, μ = v₀, v = 0; a = -2g sinθ, s = ?

Using v² - u² = 2as

⇒ (0)² - (v₀)² = 2(-2 g sinθ) s

or, s = {v₀²}/{4 g sinθ}