Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide down the same

Question

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

Accepted Answer

1. **Smooth Incline (No Friction):** The acceleration of a body sliding down a smooth incline is due to the component of gravity along the slope: $a_1 = g \sin 	heta$. The time $t_1$ to cover distance $s$ is derived from $s = \frac{1}{2}a_1 t_1^2$, giving $t_1 = \sqrt{\frac{2s}{g \sin 	heta}}$ [Source 56, 83].
2. **Rough Incline (With Friction):** The acceleration is reduced by the opposing kinetic friction force ($f_k = \mu_k mg \cos 	heta$). The net acceleration is $a_2 = g \sin 	heta - \mu_k g \cos 	heta = g(\sin 	heta - \mu_k \cos 	heta)$ [Source 82]. The time $t_2$ is $t_2 = \sqrt{\frac{2s}{a_2}}$.
3. **Relationship Given:** The problem states $t_2 = 2t_1$. Squaring both sides:
   $$ t_2^2 = 4t_1^2 $$
   $$ \frac{2s}{a_2} = 4 \left( \frac{2s}{a_1} ight) \implies a_1 = 4a_2 $$
4. **Substitution and Calculation:**
   $$ g \sin 	heta = 4 g (\sin 	heta - \mu_k \cos 	heta) $$
   Cancel $g$ and rearrange:
   $$ \sin 	heta = 4 \sin 	heta - 4 \mu_k \cos 	heta $$
   $$ 4 \mu_k \cos 	heta = 3 \sin 	heta $$
   $$ \mu_k = \frac{3}{4} 	an 	heta $$
   Given $	heta = 45^{\circ}$, $	an 45^{\circ} = 1$.
   $$ \mu_k = 0.75 $$

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