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.
Keywords:#rough inclined surface#JEE Main 2024 Evening Q42#Laws of Motion JEE Main 2024#Friction on an Inclined Plane JEE Main 2024#inclined plane#friction#block#angle
More Laws of Motion Previous-Year Questions — Page 5
A ball of mass 0.5~kg$0.5\mathrm{~kg}$ is attached to a string of length 50~cm$50\mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400~N$400\mathrm{~N}$. The maximum possible value of angular velocity of the ball in rad/s$\mathrm{rad/s}$ is:
A. 1600
B. 40
C. 1000
D. 20
Solution
Related Formula
Centripetal force configuration for simplified horizontal rotation layout:
T = mω² l$T = m\omega^2 l$
Core Logic
Given values:
m = 0.5~kg$m = 0.5\mathrm{~kg}$, l = 50~cm = 0.5~m$l = 50\mathrm{~cm} = 0.5\mathrm{~m}$, Tmax = 400~N$T_{\text{max}} = 400\mathrm{~N}$.
Ensure units are metric (50~cm → 0.5~m$50\mathrm{~cm} \to 0.5\mathrm{~m}$). Direct mapping to horizontal string projection metrics.
Chapter Mix
Class 11 Physics: Laws of Motion
Q36jee_main_2024_29_january_eveningCircular Motion and Tension
A stone of mass 900 g$900\text{ g}$ is tied to a string and moved in a vertical circle of radius 1 m$1\text{ m}$ making 10 rpm$10\text{ rpm}$. The tension in the string, when the stone is at the lowest point is (if π² = 9.8$\pi^2 = 9.8$ and g = 9.8 m/s²$g = 9.8\text{ m/s}^2$):
A.97 N$97\text{ N}$
B.9.8 N$9.8\text{ N}$
C.8.82 N$8.82\text{ N}$
D.17.8 N$17.8\text{ N}$
Solution
Related Formula
At the lowest point of a vertical circle, the equation of motion for a mass m$m$ is:
T - mg = m r ω²$$T - mg = m r \omega^2$$
Rearranging to solve for tension T$T$:
T = mg + m r ω²$$T = mg + m r \omega^2$$
Core Logic
Given data:
Mass, m = 900 g = 0.9 kg$m = 900\text{ g} = 0.9\text{ kg}$
Free-body diagram of stone at lowest point in vertical circle for Q36
Pattern Recognition
Always convert Mass to kg$\text{kg}$ and rotational speed to rad/s$\text{rad/s}$ first. Using the prompt constraint π² = 9.8$\pi^2 = 9.8$ yields a perfect decimal addition match.
A particle is moving in a circle of radius 50 cm$50\text{ cm}$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0$t = 0$ is 4 m/s$4\text{ m/s}$, the time taken to complete the first revolution will be (1)/(α)[ 1 - e-2π ] s$\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, where α =$\alpha =$ ________.
Numerical Answer.Answer: 8 to 8
Solution
Related Formula
For a particle in circular motion:
Normal (centripetal) acceleration: ac = (v²)/(r)$a_c = \frac{v^2}{r}$
Comparing this to (1)/(α)[ 1 - e-2π ] s$\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, we get:
α = 8$\alpha = 8$
Pattern Recognition
The condition aₜ = ac v (dv)/(ds) = (v²)/(r) (dv)/(v) = (ds)/(r)$a_t = a_c \implies v \frac{dv}{ds} = \frac{v^2}{r} \implies \frac{dv}{v} = \frac{ds}{r}$. Integrating directly gives v = v₀ es/r$v = v_0 e^{s/r}$. Substituting this back into v = ds/dt$v = ds/dt$ makes the final time integral much more intuitive.
Chapter Mix
Class 11 Physics: Laws of Motion
Q36jee_main_2024_27_jan_morningBanking of Tracks
A train is moving with a speed of 12 m/s$12\text{ m/s}$ on rails which are 1.5 m$1.5\text{ m}$ apart. To negotiate a curve of radius 400 m$400\text{ m}$, the height by which the outer rail should be raised with respect to the inner rail is (Given, g = 10 m/s²$g = 10\text{ m/s}^{2}$):
A.6.0 cm$6.0\text{ cm}$
B.5.4 cm$5.4\text{ cm}$
C.4.8 cm$4.8\text{ cm}$
D.4.2 cm$4.2\text{ cm}$
Solution
Related Formula
θ = (v²)/(Rg)$$\tan\theta = \frac{v^2}{Rg}$$
For small angles, θ ≈ θ = (h)/(d)$\tan\theta \approx \sin\theta = \frac{h}{d}$, where h$h$ is the raised height and d$d$ is the separation between the tracks.
Whenever θ$\theta$ is small, geometry permits approximating θ$\tan\theta$ with hwidth$\frac{h}{\text{width}}$, vastly reducing computational transcendental overhead.
Chapter Mix
Class 11 Physics: Laws of Motion
Q45jee_main_2024_27_jan_morningConservation of Linear Momentum
A body of mass 1000 kg$1000\text{ kg}$ is moving horizontally with a velocity 6 m/s$6\text{ m/s}$. If 200 kg$200\text{ kg}$ extra mass is added, the final velocity (in m/s) is:
A.6$6$
B.2$2$
C.3$3$
D.5$5$
Solution
Related Formula
m₁ v₁ = m₂ v₂$$m_1 v_1 = m_2 v_2$$
Core Logic
Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
