Starting from rest ,a body slides down a 45° inclined plane in twice the time takes to slide down the same distance in the absence of friction .The coefficient of friction between the body and inclined plane is

Sol:

Case: 1

When there is no friction:

From newton’s second law, Force = mass * acceleration.

[latex]mg sin \Theta = ma_{1}[/latex] [latex]a_{1} = g sin \Theta [/latex]

As the body starts from rest, the initial velocity (u) = 0 m/s.

Let, body travels ‘x’ distance in [latex]t_{1} time.[/latex]

From equation of motion : x = ut +[latex] \frac{1}{2} at^{2}[/latex] [latex]\Rightarrow x =\frac{1}{2} a_{1}t_{1}^{2}[/latex] [latex]\Rightarrow x = \sqrt{\frac{2x}{a_{1}}} = \sqrt{\frac{2x}{g sin \Theta }} [/latex]

Case: 2

When there is friction:

From newton’s second law, Force = mass * acceleration.

[latex]\Rightarrow mg sin\Theta – \mu mg cos \Theta = ma_{2}[/latex] [latex]\Rightarrow a_{2} = g sin \Theta – \mu g cos \Theta[/latex]

As the body starts from rest, the initial velocity (u) = 0 m/s.

Let, body travels ‘x’ distance in [latex]t_{2} time.[/latex]

From equation of motion: [latex]x = ut + \frac{1}{2} at^{2}[/latex] [latex]x = ut + \frac{1}{2} at^{2}[/latex] [latex]x = \frac{1}{2} a_{2}t_{2}^{2}[/latex] [latex]\Rightarrow t_{2} = \sqrt{\frac{2x}{a_{1}}} =\sqrt{ \frac{2x}{g sin\Theta – \mu g cos\theta}}[/latex]

It is given that t_{2} = 2t_{1}[/latex] [latex]\sqrt{\frac{2x}{g sin \theta- \mu g cos\theta}} = 2\sqrt{\frac{2x}{g {sin\theta}}}[/latex]

Now, substituting θ = 45 in the above equation, we get,

[latex]\sqrt{\frac{1}{1 – \mu }} = 2[/latex] [latex]\mu = \frac{3}{4} = 0.75[/latex]

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