Friction - JEE Advanced Previous Year Questions with Solutions

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Q. A block of mass m is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ m and tan$\theta$ >m. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from $P_{1}$ = mg (sin$\theta$ – $\mu \mathrm{cos} \theta$) to $P_{2}$=mg$(\sin \theta+\mu \cos \theta)$, the frictional force f versus P graph will look like [IIT-JEE-2010]

Ans. (A)

Q. A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $\mathrm{N}=10 \mu$, then N is [IIT-JEE-2011]

Ans. 5 $\mathrm{F}=\operatorname{mg} \sin \theta+\mu \operatorname{mg} \cos \theta$ $\mathrm{F}+\mu \mathrm{mg} \cos \theta=\operatorname{mg} \sin \theta$ $\mathrm{F}=\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta$ $\operatorname{mg} \sin \theta+\mu \operatorname{mg} \cos \theta=3 \operatorname{mg} \sin \theta-3 \mu \operatorname{mg} \cos \theta$ $4 \mu \mathrm{mg} \cos \theta=2 \mathrm{mg} \sin \mu$ $4 u \cos \theta=2 \sin \theta$ $4 \mu=2 \tan \theta$ $\mathrm{N}=10 \times \frac{1}{2}$ $\mathrm{N}=5 \quad \mu=\frac{1}{2}$

Q. A block of mass $\mathrm{m}_{1}$ = 1 kg another mass $\mathrm{m}_{2}$ = 2kg, are placed together (see figure) on an inclined plane with angle of inclination $\theta$. Various values of $\theta$ are given in List I. The coefficient of friction between the block $\mathrm{m}_{1}$ and the plane is always zero. The coefficient of static and dynamic friction between the block $\mathrm{m}_{2}$ and the plane are equal to $\mu=$ = 0.3. In List II expressions for the friction on block m2 are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g. [useful information: $\left.\tan \left(5.5^{\circ}\right) \approx 0.1 ; \tan \left(11.5^{\circ}\right) \approx 0.2 ; \tan \left(16.5^{\circ}\right) \approx 0.3\right]$ [IIT-JEE-2014]

Ans. (D) The system slip down if $\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g} \sin \theta>\mu \mathrm{m}_{2} \mathrm{gcos} \theta$ $\tan \theta>\frac{\mu \mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}>\frac{0.3 \times 2}{3}$ $\tan \theta>0.2$ $\Rightarrow \theta>11.5^{\circ}$ For $\mathrm{P}$ and $\mathrm{Q}$ system will remain stationary hence friction $=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g} \sin \theta$ For $\mathrm{R}$ and $\mathrm{S}$ system will move hence limiting friction acts friction $=\mu \mathrm{m}_{2} \mathrm{g} \cos \theta$

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Friction problems are an important part of Mechanics and often require careful application of concepts of force, equilibrium, and motion. Students looking for additional Irodov problems for JEE Advanced can refer to Irodov Selected Questions, offering exposure to challenging Physics problems across different concepts.