Find the coefficient of friction if the time taken by a block on the rough surface is \(50\%\) more than the time taken on a smooth surface. The distance slid by the mass is the same in both cases. The incline angle is \(45^\circ\).
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When distance travelled is the same for motions starting from rest, use
\( s = \tfrac{1}{2}at^2 \).
Thus \( a_1 t_1^2 = a_2 t_2^2 \).
For an inclined plane with friction, acceleration is \( g(\sin\theta - \mu\cos\theta) \).
To solve this problem, we need to determine the coefficient of friction (\( \mu \)) given that the time taken by a block on a rough surface is \(50\%\) more than the time on a smooth surface. The distance is the same, and the incline angle is \(45^\circ\).
Let's denote:
\( t_1 \) as the time taken on the smooth surface.
\( t_2 \) as the time taken on the rough surface, which is \(1.5 \times t_1\).
\( s \) as the distance slid, same for both surfaces.
On a smooth surface, the only force causing acceleration is the component of gravity along the incline. Therefore, the acceleration \( a_1 \) is given by: \(a_1 = g \sin \theta\) where \( \theta = 45^\circ \), so \( \sin 45^\circ = \frac{\sqrt{2}}{2} \).
On a rough surface, friction also acts along with gravity, slowing down the block. Thus, the acceleration \( a_2 \) is: \(a_2 = g \sin \theta - \mu g \cos \theta\).
Using the equation of motion for both surfaces:
For the smooth surface: \(s = \frac{1}{2} a_1 t_1^2\) Substituting \( a_1 = g \sin \theta \): \(s = \frac{1}{2} g \sin \theta t_1^2\).
For the rough surface: \(s = \frac{1}{2} a_2 t_2^2\) Substituting \( a_2 = g \sin \theta - \mu g \cos \theta \): \(s = \frac{1}{2} \left( g \sin \theta - \mu g \cos \theta \right) (1.5 t_1)^2\).
Since the distance \( s \) is the same in both cases, equate the above two equations: \(\frac{1}{2} g \sin \theta t_1^2 = \frac{1}{2} \left( g \sin \theta - \mu g \cos \theta \right) (1.5 t_1)^2\).
