In the given arrangement of a doubly inclined plane two blocks of masses M$M$ and m$m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25$0.25$. The value of m$m$, for which M = 10mathrm\ kg$M = 10\mathrm{\ kg}$ will move down with an acceleration of 2mathrm\ m/s^2$2\mathrm{\ m/s^2}$ is : (take g = 10mathrm\ m/s^2$g = 10\mathrm{\ m/s^2}$ and tan 37^circ = 3 / 4$\tan 37^{\circ} = 3 / 4$)

### Related Formula f_s leq mu_s N$$f_s \leq \mu_s N$$ F_c = mromega^2$$F_c = mr\omega^2$$ ### Core Logic

To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. f = momega^2 r$f = m\omega^2 r$ The normal force on the flat disc is N = mg$N = mg$. The maximum static friction is f_textmax = mu N = mu mg$f_{\text{max}} = \mu N = \mu mg$. For no slipping: m r omega^2 leq mu mg$$m r \omega^2 \leq \mu mg$$ omega^2 leq fracmu gr$$\omega^2 \leq \frac{\mu g}{r}$$ omega_textmax = sqrtfracmu gr$$\omega_{\text{max}} = \sqrt{\frac{\mu g}{r}}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws Of Motion