A body slides down a rough plane inclined to the horizontal at 30^0…

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A body slides down a rough plane inclined to the horizontal at $30^{\circ}$ . If $70\%$ of the initial potential energy is dissipated during the descent, find the coefficient of sliding friction.

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2 Explanation of Terms:

4 Final Answer:

Explanation:

When an object slides down an inclined rough plane, different forces act on it:

Gravitational force ( $mg$ ): Weight of the object which acts downward.
Normal force ( $N$ ): Force exerted on the object by the surface perpendicular to the surface.
Frictional force ( $f$ ): Force produced due to friction between the surface and the object, which acts in the opposite direction to the motion.

Due to friction, some energy of the object is damped. According to the principle of energy conservation, some part of the initial potential energy is converted into work of friction.

Explanation of Terms:

$m$ : Mass of the object
$g$ : Acceleration due to gravity ( $9.8 \, \text{m/s}^2$ )
$h$ : Height, $h = s \sin \theta$
$s$ : Length of inclined plane
$\theta$ : Angle of inclination ( $30^\circ$ )
$\mu$ : Coefficient of friction
$N$ : Normal force, $N = mg \cos \theta$
$W_{\text{friction}}$ : Work done due to friction
$\text{PE}_{\text{initial}}$ : Initial potential energy

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Solution:

Initial potential energy ( $\text{PE}_{\text{initial}}$ ):

$\text{PE}_{\text{initial}} = mgh = mg s \sin \theta$

Work done due to friction ( $W_{\text{friction}}$ ):

$W_{\text{friction}} = -f s = -\mu N s$

Where normal force $N$ is:

$N = mg \cos \theta$

So,

$W_{\text{friction}} = -\mu mg \cos \theta \, s$

Dampened energy:

If $70\%$ of the energy is damped due to friction, then:

$W_{\text{friction}} = -0.70 \times \text{PE}_{\text{initial}}$

Set up the equation Do:

$-\mu mg \cos \theta \, s = -0.70 \times mg s \sin \theta$

Simplify:

Subtract $mg s$ from both sides:

$-\mu \cos \theta = -0.70 \sin \theta$

$\mu \cos \theta = 0.70 \sin \theta$

Find the coefficient of friction ( $\mu$ ):

$\mu = \frac{0.70 \sin \theta}{\cos \theta}$

$\mu = 0.70 \tan \theta$

Since $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.5774$ , then:

$\mu = 0.70 \times 0.5774$

$\mu \approx 0.4042$

Final Answer:

The coefficient of friction ( $\mu$ ) is approximately $\bf 0.404$ .

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