Question

Consider three fixed surfaces shown in the figure. Three blocks each of mass m are released from rest from top of the three surfaces. All blocks reach ground with same speed. Length of path travelled by the blocks is same for second and third surface. If coefficient of friction of three surfaces are $\mu_1$, $\mu_2$ and $\mu_3$ respectively then

• A. $\mu_1$ = $\mu_2$
• B. $\mu_1 > \mu_2$
• C. $\mu_2 = \mu_3$
• D. $\mu_2 < \mu_3$

Answer 1

Answer: (B), (D)

(B) and (D)

Answer 2

By the work-energy theorem, the work done by friction is the same for all three surfaces since the initial and final kinetic energies, as well as the initial potential energy, are the same.

For surface 1 (straight incline), the work done by friction is $W_{f1} = -\mu_1 N_1 L_1 = -\mu_1 (mg \cos\theta_1) L_1 = -\mu_1 mgl$.

For surfaces 2 and 3, the path length is $L$. $W_{f2} = -\mu_2 \int N_2 ds = -\mu_2 \langle N_2 \rangle L$ $W_{f3} = -\mu_3 \int N_3 ds = -\mu_3 \langle N_3 \rangle L$

Since $W_{f1} = W_{f2} = W_{f3}$, we have $\mu_1 mgl = \mu_2 \langle N_2 \rangle L = \mu_3 \langle N_3 \rangle L$.

For surface 2 (concave), the normal force is greater than the perpendicular component of gravity ($\langle N_2 \rangle > \langle N_1 \rangle$). For surface 3 (convex), the normal force is less than the perpendicular component of gravity ($\langle N_3 \rangle < \langle N_1 \rangle$).

Since $\mu_1 \langle N_1 \rangle L_1 = \mu_2 \langle N_2 \rangle L$ and $L \approx L_1$, and $\langle N_2 \rangle > \langle N_1 \rangle$, it implies $\mu_1 > \mu_2$.

Since $\mu_1 \langle N_1 \rangle L_1 = \mu_3 \langle N_3 \rangle L$ and $L \approx L_1$, and $\langle N_3 \rangle < \langle N_1 \rangle$, it implies $\mu_1 < \mu_3$.

Comparing surfaces 2 and 3: $\mu_2 \langle N_2 \rangle L = \mu_3 \langle N_3 \rangle L \implies \mu_2 \langle N_2 \rangle = \mu_3 \langle N_3 \rangle$. Since $\langle N_2 \rangle > \langle N_3 \rangle$, it implies $\mu_2 < \mu_3$.