Question

a box slides down at constant velocity along an inclined plane which makes angle x with the horizon. if the box is now given an initial velocity v in the upward direction along the plane, what would be the displacement of the box

Answer 1

Answer on Question #56323, Physics Mechanics Relativity

a box slides down at constant velocity along an inclined plane which makes angle x with the horizon. if the box is now given an initial velocity v in the upward direction along the plane, what would be the displacement of the box

Solution

Fig.1

From Newton's second law

$$m \vec {g} + \hat {f} _ {k} + \vec {N} = 0$$

$m$ is the mass of box; $g$ is the gravity acceleration; $N$ is reaction force; $f_{k} = \mu N$ is friction force.

Then

$$\left\{ \begin{array}{l} f _ {k} = m g \sin \alpha \\ N = m g \cos \alpha \end{array} \right.$$

So,

$$m g \sin \alpha = \mu m g \cos \alpha \Rightarrow \mu = t g \alpha$$

The time of motion

$$t = \frac {v}{g \mu \cos \alpha + g \sin \alpha} = \frac {v}{g \cdot t g \alpha \cos \alpha + g \cdot \sin \alpha} = \frac {v}{g \cos \alpha \cdot \frac {\sin \alpha}{\cos \alpha} + g \cdot \sin \alpha} = \frac {v}{2 g \cdot \sin \alpha}$$

be the displacement of the box

$$l = \frac {v ^ {2}}{2 g (\mu \cos \alpha + \sin \alpha)} = \frac {v ^ {2}}{2 g (t g \alpha \cdot \cos \alpha + \sin \alpha)} = \frac {v ^ {2}}{4 g \sin \alpha}$$

Answer: $l = \frac{v^2}{4g\sin\alpha}$

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