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Standard XII

Physics

Friction General Understanding

The coefficie...

Question

The coefficient of friction between a body and the surface of an inclined plane at $45^{o}$ is $0.5$ . If $g = 9.8 {m / s}^{2}$ the acceleration of the body downward is ${m / s}^{2}$ is

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Solution

If body is going up, then
$μ_{d o w n w a r d} = \frac{m g s i n 45 + μ m g c o s 45}{m} g (\frac{1}{\sqrt{2}} + \frac{0.5}{\sqrt{2}}) g (\frac{1}{\sqrt{2}} + \frac{1}{2 \sqrt{2}}) = g (\frac{3}{2 \sqrt{2}}) = 4.9 (\frac{3}{\sqrt{2}}) = 10.4 m / s^{2}$
If body is going down, then
$a_{d o w n} = g s i n 45 = 9.8 \times \frac{1}{\sqrt{2}} = 6.92 m / s^{2}$

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The coefficient of friction between a body and the surface of an inclined plane at $45^{o}$ is 0.5. If g=9.8 $m / s^{2}$ the acceleration of the body downwards in $m / s^{2}$ is

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The coefficient of friction between a body and the surface of an inclined plane at 45o is 0.5. If g=9.8 m/s2 the acceleration of the body downward is m/s2 is

If body is going up, then μdownward=mgsin45+μmgcos45mg(1√2+0.5√2)g(1√2+12√2)=g(32√2)=4.9(3√2)=10.4m/s2If body is going down, then adown=gsin45=9.8×1√2=6.92m/s2

The coefficient of friction between a body and the surface of an inclined plane at $45^{o}$ is $0.5$ . If $g = 9.8 {m / s}^{2}$ the acceleration of the body downward is ${m / s}^{2}$ is