Question

A body is sliding down a rough inclined plane which makes an angle of 30 degree with the horizontal. If the coefficient of friction is 0.26, Determine the acceleration in ms-2.​

Answer 1

Answer:

a= 9.8

( 1/2- 0.3 ×√3/2 )

= 2.35ms²

Answer 2

Concept:

• First draw the Free Body Diagram of the object sliding.
• component of mg will responsible for motioning it down.
• And friction will oppose it.

$\implies\rm F_{net} = F_{driving}-F_{opposing} = ma$

$\implies \rm Friction =\mu N = \mu mgcos\theta$

Solution:

$\to\rm mgsin\theta - \mu mgcos\theta = ma$

$\to\rm gsin30\degree - \mu gcos30\degree = a$

$\to\rm g(sin30 - \mu cos30) = a$

$\to\rm g\bigg(\dfrac{1}{2}- 0.26 \times \dfrac{\sqrt{3}}{2}\bigg) = a$

$\to\rm g(0.5 - 0.22) = a$

$\to\rm a = 10 \times 0.28$

$\to\bf a \approx 2.8 ms^{-2}$

★ so acceleration for sliding down is approximately 2.8 m/s²

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