Question

$\boxed{\LARGE{\sf \maltese Physics \:Question:-}}$

A block of mass 8kg is sliding on a surface inclined at an angle of 45° with the horizontal. Calculate the acceleration of the block.The coefficient of friction between the block and the surface is 0.6.(take g=10m/s${}^2$


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Answer 1

Need to FinD :-

• The acceleration of the block .

$\red{\frak{Given}}\begin{cases}\sf Mass \ of \ block \ = \ 8kg .\\\\\sf Angle_{( of\ inclined\ surface )}= 45^o \\\\\sf Coefficient \ of \ friction \ = 0.6 \\\\\sf Accl^n \ due \ to \ gravity \ = \ 10m/s^2 \end{cases}$

Given that the mass of the block is 8kg and it is sliding on a surface inclined at angle 45° . For the figure refer to attachment.

• Gravitational force , will be a acting vertically downwards , i.e. mg .
• The plane is inclined at an angle of 45° .On breaking the components of mg, we get , mg sin 45° acting along the slope of the plane and mg cos 45° perpendicular to the surface .
• Frictional force will be acting in direction opposite to mg sin45° .
• Normal force will be acting perpendicular to the surface of plane.

The component of force acting along the surface of inclined plane is , mg sin45° .

$\rule{200}4$

We know ,

$\sf\small \longrightarrow Force_{(Friction)}= \mu \ N$

• Here N is mg cos45° .

$\sf\small \longrightarrow Force_{(Friction)}= \mu \ mg \ cos45^o$

$\rule{200}4$

Therefore ,

$\sf\small \longrightarrow Force_{(net)}= mg \ sin45^o - \mu \ mg \ cos 45^o \\\\\\ \sf\small \longrightarrow mass * a_{(net)}= mg ( sin45^o - 0.6 cos 45^o) \\\\\\ \sf\small \longrightarrow mass * a_{(net)}= mg\bigg( \dfrac{1}{\sqrt2}-0.6\times \dfrac{1}{\sqrt2} \bigg) \\\\\\ \sf\small\longrightarrow mass * a_{(net)}= \dfrac{mg}{\sqrt2} ( 1 - 0.6 ) \\\\\\ \sf\small\longrightarrow 8 * a_{(net)}= \dfrac{mg}{\sqrt2} \times 0.4 \\\\\\ \sf\small\longrightarrow a_{(net)} = \dfrac{ 8 * 10 * 0.4 }{8\sqrt2} \\\\\\ \sf\small\longrightarrow\underline{\underline{\red{ a_{(net)}= 2\sqrt{2} m/s^2 \approx 2.828 \ m/s^2 }}}$

$\rule{200}4$

Answer 2

REFER TO ATTACHMENT

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