A block of mass 5 kg is moving on an inclined plane which makes an angle of 30°$30^{\circ}$ with the horizontal. Friction coefficient between the block and inclined plane surface is √(3)2$\frac{\sqrt{3}}{2}$ . The force to be applied on the block so that the block will move down without acceleration is ____ N.
A.25$25$
B.12.5$12.5$
C.7.5$7.5$
D.15$15$
Solution & Explanation
Related Formula
fk = μk N = μk mg θ$$f_k = \mu_k N = \mu_k mg \cos\theta$$Fdown = mg θ$$F_{\text{down}} = mg \sin\theta$$
Net force = 0 for constant velocity$\text{Net force } = 0 \text{ for constant velocity}$
Core Logic
For the block to move down at a constant velocity (zero acceleration), the sum of forces parallel to the incline must be zero.
Free body diagram of block on an incline
Step 1: Force Balance Equation
Force pulling it down the incline = mg (30°)$mg \sin(30^{\circ})$
Friction acting up the incline (since block moves down) = μ mg (30°)$\mu mg \cos(30^{\circ})$
Let external force F$F$ act down the incline.
mg (30°) + F = μ mg (30°)$$mg \sin(30^{\circ}) + F = \mu mg \cos(30^{\circ})$$
Note: The PDF solution calculates this with F assigned up the incline and gets F = -12.5N$F = -12.5\mathrm{N}$. The magnitude is 12.5N$12.5\mathrm{N}$ downward on the incline.
Pattern Recognition
If μ > θ$\mu > \tan\theta$ (here √(3)2 = 0.866$\frac{\sqrt{3}}{2} = 0.866$ and 30^° = 1√(3) ≈ 0.577$\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$), friction is stronger than gravity's pull. You must actively pull it down to keep it moving.
Keywords:#force to be applied on the block#JEE Main 2026 Morning Q40#Laws of Motion JEE Main 2026#Friction on an Inclined Plane JEE Main 2026
More Laws of Motion Previous-Year Questions — Page 3
Q14jee_main_2025_07_april_morningFriction
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
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.
Chapter Mix
Class 11 Physics: Laws of Motion
Q15jee_main_2025_08_april_eveningNewton's Second Law
A body of mass 2~kg$2\mathrm{~kg}$ moving with velocity of vᵢₙ = 3 i +4 j~m/s$\vec{v}_{\mathrm{in}} = 3\hat{i} +4\hat{j}\mathrm{~m/s}$ enters into a constant force field of 6~N$6\mathrm{~N}$ directed along positive z-axis. If the body remains in the field for a period of (5)/(3)$\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is:
F = m a$$\vec{F} = m \vec{a}$$v = u + at$$\vec{v} = \vec{u} + \vec{a}t$$
where,
F$\vec{F}$ = constant force vector
m$m$ = mass of body
a$\vec{a}$ = acceleration vector
u$\vec{u}$ = initial velocity vector
v$\vec{v}$ = final velocity vector
Core Logic
Given parameters:
Mass, m = 2~kg$m = 2\mathrm{~kg}$
Initial velocity, u = 3 i + 4 j~m/s$\vec{u} = 3\hat{i} + 4\hat{j}\mathrm{~m/s}$
Force, F = 6 k~N$\vec{F} = 6\hat{k}\mathrm{~N}$ (directed along positive z-axis)
Time interval, t = (5)/(3)~s$t = \frac{5}{3}\mathrm{~s}$
Calculate the acceleration vector a$\vec{a}$:
a = Fm = 6 k2 = 3 k~m/s²$$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k}\mathrm{~m/s}^{2}$$
Thus, the emerging velocity of the body is 3 i + 4 j + 5 k~m/s$3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$.
Pattern Recognition
Sees: Orthogonal initial velocity and force field direction.
Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged (3 i + 4 j$3\hat{i} + 4\hat{j}$). Simply compute the z-component change: vz = az t = ((6)/(2)) ((5)/(3)) = 5$v_z = a_z t = \left(\frac{6}{2}\right) \left(\frac{5}{3}\right) = 5$. Result: 3 i + 4 j + 5 k$3\hat{i} + 4\hat{j} + 5\hat{k}$. ✓
Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
Q12jee_main_2025_29_jan_eveningVariable Mass System
A sand dropper drops sand of massm(t)$m(t)$ on a conveyer belt at a rate proportional to the square root of speed (v$v$) of the belt, i.e. (dm)/(dt) ∝ √(v)$\frac{dm}{dt} \propto \sqrt{v}$ . If P$P$ is the power delivered to run the belt at constant speed then which of the following relationship is true?
(dm)/(dt) = C √(v) (where C is a constant)$$\frac{dm}{dt} = C \sqrt{v} \quad (\text{where } C \text{ is a constant})$$
To maintain a constant velocity v$v$, the continuous force applied by the conveyor system must balance the rate of gain of momentum of the dropped sand:
F = ((dm)/(dt)) v = (C √(v)) · v = C v3/2$$F = \left(\frac{dm}{dt}\right) v = (C \sqrt{v}) \cdot v = C v^{3/2}$$
Power delivered is the product of force and speed:
P = F · v = (C v3/2) · v = C v5/2$$P = F \cdot v = (C v^{3/2}) \cdot v = C v^{5/2}$$
Squaring both sides of the expression:
P² ∝ v⁵$P^2 \propto v^5$
Pattern Recognition
In variable mass problems involving dropping dust/sand at rest onto a moving frame, the thrust force always simplifies to v · (dm)/(dt)$v \cdot \frac{dm}{dt}$, which makes power scale as v² · (dm)/(dt)$v^2 \cdot \frac{dm}{dt}$.
Chapter Mix
Class 11 Physics: Laws of Motion
Q6jee_main_2025_03_april_morningSpring-Block Dynamics with Friction
Two blocks of masses m and M, (M > m)$(\mathbf{M} > \mathbf{m})$ , are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released then
(μ =coefficient of friction between the two blocks)$(\mu =\text{coefficient of friction between the two blocks})$The diagram displays a smaller block of mass m placed on top of a larger block of mass M, which is connected to a horizontal spring on a smooth table.
(A) The time period of small oscillation of the two blocks is T = 2π (m + M)k$\mathrm{T} = 2\pi \sqrt{\frac{(\mathrm{m} + \mathrm{M})}{\mathrm{k}}}$
(B) The acceleration of the blocks is a = (kx)/(M + m)$a = \frac{kx}{M + m}$ (x = displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is mk|x|M + m$\frac{\mathrm{m}k|\mathrm{x}|}{\mathrm{M} + \mathrm{m}}$
(D) The maximum amplitude of the upper block, if it does not slip, is μ(M + m)gk$\frac{\mu(\mathbf{M} + \mathbf{m})\mathbf{g}}{\mathbf{k}}$
(E) Maximum frictional force can be μ (M + m)g$\mu (\mathbf{M} + \mathbf{m})\mathbf{g}$.
Choose the correct answer from the options given below:
A. A, B, D Only
B. B, C, D Only
C. C, D, E Only
D. A, B, C Only
Solution
Related Formula
For combined system performing simple harmonic motion without relative slipping:
T = 2π mtotalk$$T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}}$$a = -ω² x = -(k)/(M+m) x$$a = -\omega^2 x = -\frac{k}{M+m} x$$
Core Logic
Let's analyze each statement:
Statement (A): Since both blocks perform SHM together, the combined mass is (M + m)$(M + m)$. The spring constant is k$k$. Thus, the time period of small oscillation is:
T = 2π √((M+m)/(k))$$T = 2\pi \sqrt{\frac{M+m}{k}}$$
This is correct. (A is True)
Statement (B): When the system is displaced by x$x$, the restoring spring force on the combined system is F = -kx$F = -kx$. The common acceleration of the combined mass is:
This matches the expression (taking magnitude). (B is True)
Statement (C): The upper block of mass m$m$ moves solely due to the static frictional force f$f$ acting on it. Thus:
f = m a = m ( (kx)/(M+m) ) = (mkx)/(M+m)$$f = m a = m \left( \frac{kx}{M+m} \right) = \frac{mkx}{M+m}$$
Statement (C) claims the frictional force is (mμ|x|)/(M+m)$\frac{m\mu|x|}{M+m}$, which is incorrect because friction is determined by acceleration, not by coefficient of friction μ$\mu$ during static grip. (C is False)
Statement (D): For no slipping to occur, the maximum frictional force required at peak amplitude A$A$ must be less than or equal to the limiting static friction fL = μ mg$f_L = \mu mg$:
fmax = (mkA)/(M+m) ≤ μ mg$$f_{\text{max}} = \frac{mkA}{M+m} \le \mu mg$$(kA)/(M+m) ≤ μ g A ≤ (μ(M+m)g)/(k)$$\frac{kA}{M+m} \le \mu g \implies A \le \frac{\mu(M+m)g}{k}$$
Thus, the maximum amplitude is (μ(M+m)g)/(k)$\frac{\mu(M+m)g}{k}$. (D is True)
Statement (E): The maximum static frictional force between the blocks is fL = μ mg$f_L = \mu mg$, not μ (M+m)g$\mu (M+m)g$. (E is False)
Step 1: Conclusion
Only statements A, B, and D are correct. Hence, the correct option is (1).
The diagram displays a smaller block of mass m placed on top of a larger block of mass M, which is connected to a horizontal spring on a smooth table.The diagram displays a smaller block of mass m placed on top of a larger block of mass M, which is connected to a horizontal spring on a smooth table.
Pattern Recognition
In stacked blocks with springs, always identify the force driving the non-spring-loaded block. Here, mass m$m$ is driven purely by friction, so f = m · a$f = m \cdot a$. Slipping begins when this required force exceeds flimit = μ m g$f_{\text{limit}} = \mu m g$. This simple boundary matches the derivation of maximum amplitude perfectly!
Chapter Mix
Class 11 Physics: Laws of Motion: Friction
Class 11 Physics: Oscillations: Simple Harmonic Motion
Q14jee_main_2025_04_april_eveningFriction
A block of mass 25 kg is pulled along a horizontal surface by a force at an angle 45circ$45^{circ}$ with the horizontal. The friction coefficient between the block and the surface is 0.25. The work done for a displacement of 5 m of the block with uniform velocity is:
A. 970 J
B. 735 J
C. 245 J
D. 490 J
Solution
Related Formula
N + F θ = mg N = mg - F θ$$N + F\sin\theta = mg \implies N = mg - F\sin\theta$$F θ = fk = μk N$$F\cos\theta = f_k = \mu_k N$$W = F · S · θ$$W = F \cdot S \cdot \cos\theta$$
Core Logic
Since the block moves with uniform velocity, horizontal acceleration is zero.
F√(2) = (61.25)/(1.25) = 49 F = 49√(2) N$$\frac{F}{\sqrt{2}} = \frac{61.25}{1.25} = 49 \implies F = 49\sqrt{2}\text{ N}$$
The work done by the external force over displacement S=5 m$S=5\text{ m}$ is:
W = F S (45^°) = (49√(2)) × 5 × 1√(2) = 245 J$$W = F S \cos(45^\circ) = (49\sqrt{2}) \times 5 \times \frac{1}{\sqrt{2}} = 245\text{ J}$$Free body diagram of the block showing external force components and friction
Pattern Recognition
When uniform velocity is sustained, work done by the external pulling component perfectly matches the work consumed against internal friction dissipation (W = fk · S$W = f_k \cdot S$).
Chapter Mix
Class 11 Physics: Laws of Motion
More Laws of Motion Questions — jee_main_2026_28_january_morning
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.
