Question

a piece of ice slides down at 45 degree incline plane in twice the time it takes to slide down a frictionless 45 degree inclined plane. what is the coefficient of friction between the ice and the incline plane?

Answer 1

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➡If the time is doubled by the presence of friction, the acceleration is reduced by a factor 4. (that is because S = a × t^2 / 2 remains the same).

➡The friction is therefore 3/4 of the component of the weight down the  incline.

=> m × g × cos 45 × k = (3/4) × m × g × sin 45, where k is the  kinetic friction coefficient

..k = 3/4 => 0.75...Ans..✔

Answer 2

Answer:

The answer is 0.75.

Explanation:

In both the point, the initial speed u and the distance s traveled exists

According to the formula: $s=u t+\frac{1}{2} a t^{2}$ exists equal (as u=0 in both the cases).

Let a1 be the acceleration of the ice block on a rough incline and t1 be the time brought to slide down this rough plane. Also, let a2 and t2 be equal quantities in the case of the frictionless inclined plane.

Given that $\mathrm{t}_{1}=2 \mathrm{t}_{2}$

Further, $\frac{a_{1}}{a_{2}}=\frac{t_{2}^{2}}{t_{1}^{2}}\left(\because a^{2}=\right.constant)$

$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{t}_{2}^{2}}{4 \mathrm{t}_{2}^{2}}=\frac{1}{4} \ldots$  (1)

But $a_{1}=g(\sin \theta-\mu \cos \theta) \ldots$(2)

and $a_{2}=g \sin \theta \ldots$ (3)

$\therefore \frac{g(\sin \theta-\mu \cos \theta)}{g \sin \theta}=\frac{1}{4}$

$\Rightarrow 4 \sin \theta-4 \mu \cos \theta=\sin \theta$

$\Rightarrow 3 \sin \theta=4 \mu \cos \theta$

$\therefore \mu=\frac{3}{4} \tan \theta=\\\frac{3}{4} \tan 45^{\circ} \Rightarrow \mu=\frac{3}{4}=0.75$

Therefore, the answer is 0.75.

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