A cubic block of mass m is sliding down on an inclined plane at 60°$60^{\circ}$ with an acceleration of (g)/(2)$\frac{g}{2}$ , the value of coefficient of kinetic friction is
A.√(3) - 1$\sqrt{3} - 1$
B.√(3)2$\frac{\sqrt{3}}{2}$
C.√(2)3$\frac{\sqrt{2}}{3}$
D.1 - √(3)2$1 - \frac{\sqrt{3}}{2}$
Solution & Explanation
Related Formula
For a block sliding down an inclined plane of inclination θ$\theta$:
mg θ - fk = ma$$mg \sin\theta - f_k = ma$$
Where normal reaction is N = mg θ$N = mg \cos\theta$ and kinetic friction is:
fk = μk N = μk mg θ$$f_k = \mu_k N = \mu_k mg \cos\theta$$
Sees: Block sliding down with acceleration on an incline.
Shortcut: The acceleration on an incline is a = g( θ - μk θ)$a = g(\sin\theta - \mu_k \cos\theta)$. For θ = 60^°$\theta = 60^\circ$, this is a = g( √(3)2 - (μk)/(2))$a = g\left(\frac{\sqrt{3}}{2} - \frac{\mu_k}{2}\right)$. Equating to g/2$g/2$ gives μk = sqrt3 - 1$\mu_k = sqrt{3} - 1$ directly.
Keywords:#coefficient of kinetic friction on incline#JEE Main 2025 Morning Q14#Laws of Motion JEE Main 2025#sliding block incline mechanics
More Laws of Motion Previous-Year Questions — Page 7
Q31jee_main_2024_31_jan_eveningPulley Systems and Tension
A light string passing over a smooth light fixed pulley connects two blocks of masses m₁$m_1$ and m₂$m_2$. If the acceleration of the system is g/8$g/8$, then the ratio of masses is
The image displays two masses suspended over a single fixed smooth pulley.
For standard Atwood machines, a = g × Difference in massSum of mass$a = g \times \frac{\text{Difference in mass}}{\text{Sum of mass}}$. If a/g = 1/8$a/g = 1/8$, then (m₁-m₂)/(m₁+m₂) = 1/8$(m_1-m_2)/(m_1+m_2) = 1/8$, which can be solved using componendo and dividendo: m₁/m₂ = (8+1)/(8-1) = 9/7$m_1/m_2 = (8+1)/(8-1) = 9/7$.
Chapter Mix
Class 11 Physics: Laws of Motion
Q42jee_main_2024_31_jan_eveningFriction on an Inclined Plane
A block of mass 5 kg$5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
If F₁$\vec{F}_1$ is the force required to just move the block up the inclined plane and F₂$\vec{F}_2$ is the force required to just prevent the block from sliding down, then the value of | F₁| - | F₂|$|\vec{F}_1| - |\vec{F}_2|$ is: [Use g = 10 m/s²$g = 10 \text{ m/s}^2$]
To move the block up, the applied force F₁$F_1$ must overcome both the downward gravitational component and the downward frictional force.
To prevent it from sliding down, the applied force F₂$F_2$ acts upwards and is aided by friction which acts upwards to oppose impending downward slip.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
The image shows a 5 kg block on a rough plane inclined at 30 degrees, with coefficient of friction mu = 0.1.
Note: The official options had an anomaly where 5√(3) N$5\sqrt{3} \text{ N}$ was missing or evaluated as a bonus. Option 2 was listed as 50√(3)$50\sqrt{3}$ in the primary text. We track the closest logic path indicating Bonus.
Pattern Recognition
The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely 2 fk$2 f_k$ (2 μ mg θ$2 \mu mg \cos \theta$). Bypass calculating the mg θ$mg \sin \theta$ terms entirely.
Chapter Mix
Class 11 Physics: Laws of Motion
Qjee_main_2024_31_jan_morningPulley And Incline Friction
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 = 10 kg$M = 10\mathrm{\ kg}$ will move down with an acceleration of 2 m/s²$2\mathrm{\ m/s^2}$ is : (take g = 10 m/s²$g = 10\mathrm{\ m/s^2}$ and 37° = 3 / 4$\tan 37^{\circ} = 3 / 4$)
Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.
A.9 kg$9\mathrm{\ kg}$
B.4.5 kg$4.5\mathrm{\ kg}$
C.6.5 kg$6.5\mathrm{\ kg}$
D.2.25 kg$2.25\mathrm{\ kg}$
Solution
Related Formula
Σ F = ma$$\sum F = ma$$fk = μk N = μk mg θ$$f_k = \mu_k N = \mu_k mg \cos\theta$$
Core Logic
Two blocks M and m on opposite sides of a double inclined plane linked by a rope over a top pulley. M is on the 53-degree slope and moving downwards, m is on the 37-degree slope.
Since block M$M$ moves down the incline, kinetic friction opposes its motion (acts upwards).
Block m$m$ is pulled up the incline, so kinetic friction opposes its motion (acts downwards).
A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ$\mu$. If the distance of the coin from the center of the disc is r$r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
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