A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent the block from sliding down, then the value of |F1| - |F2| is: [Use g = 10 m/s2]

To move the block up, the applied force F1 must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force F2 acts upwards and is aided by friction which acts upwards to oppose impending downward slip. Moving up: Preventing slip down: *Note: The official options had an anomaly where 5sqrt(3) N was missing or evaluated as a bonus. Option 2 was listed as 50sqrt(3) in the primary text. We track the closest logic path indicating Bonus.* The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely 2 fk (2 mu mg theta). Bypass calculating the mg theta terms entirely. Class 11 Physics: Laws of Motion

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent

To move the block up, the applied force F1 must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force F2 acts upwards and is aided by friction which acts upwards to oppose impending downward slip. Moving up: Preventing slip down: *Note: The official options had an anomaly where 5sqrt(3) N was missing or evaluated as a bonus. Option 2 was listed as 50sqrt(3) in the primary text. We track the closest logic path indi

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent the block from sliding down, then the value of |F1| - |F2| is: [Use g = 10 m/s2]

To move the block up, the applied force F1 must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force F2 acts upwards and is aided by friction which acts upwards to oppose impending downward slip. Moving up: Preventing slip down: *Note: The official options had an anomaly where 5sqrt(3) N was missing or evaluated as a bonus. Option 2 was listed as 50sqrt(3) in the primary text. We track the closest logic path indicating Bonus.* The difference between 'push up' and 'hold from sliding' forces on an incline is always precisely 2 fk (2 mu mg theta). Bypass calculating the mg theta terms entirely. Class 11 Physics: Laws of Motion

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent

To move the block up, the applied force F1 must overcome both the downward gravitational component and the downward frictional force. To prevent it from sliding down, the applied force F2 acts upwards and is aided by friction which acts upwards to oppose impending downward slip. Moving up: Preventing slip down: *Note: The official options had an anomaly where 5sqrt(3) N was missing or evaluated as a bonus. Option 2 was listed as 50sqrt(3) in the primary text. We track the closest logic path indi

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just…

Year	2026	2025	2024	Total
Questions	9	10	15	34

Q31 jee_main_2026_21_jan_morning Force and Acceleration

A 4 kg mass moves under the influence of a force

F = (4t³ i - 3t j) N

where t is the time in second. If mass starts from origin at

t = 0

, the velocity and position after

t = 2s

will be:

A. $v = 3 i +(3)/(2) j, r = (6)/(5) i + j$
B. $v = 4 i -(3)/(2) j, r = (8)/(5) i - j$
C. $v = 4 i +(5)/(2) j, r = (8)/(5) i +2 j$
D. $v = 4 i -(3)/(2) j, r = (6)/(5) i - j$

Solution

Related Formula

a = Fm

v = \int a dt

r = \int v dt

Core Logic

Given $F = 4t³ i - 3t j$ $\vec{F} = 4t^3\hat{i} - 3t\hat{j}$ and mass $m = 4 kg$ $m = 4\text{ kg}$. Acceleration:

a = Fm = 4t³ i - 3t j4 = t³ i - (3)/(4)t j

Step 1: Calculating Velocity

Velocity is the integral of acceleration with $v(0) = 0$ $v(0) = 0$:

v = \int ( t³ i - (3)/(4)t j ) dt = (t⁴)/(4) i - (3)/(8)t² j

At $t = 2 s$ $t = 2\text{ s}$:

v(2) = ((2)⁴)/(4) i - (3)/(8)(2)² j = 4 i - (3)/(2) j

Step 2: Calculating Position

Position is the integral of velocity with $r(0) = 0$ $\vec{r}(0) = \vec{0}$:

r = \int v dt = \int ( (t⁴)/(4) i - (3)/(8)t² j ) dt

r = (t⁵)/(20) i - (3)/(24)t³ j = (t⁵)/(20) i - (t³)/(8) j

At $t = 2 s$ $t = 2\text{ s}$:

r(2) = ((2)⁵)/(20) i - ((2)³)/(8) j = (32)/(20) i - (8)/(8) j = (8)/(5) i - j

Pattern Recognition

Recognize $F \to a \to v \to r$ $F \to a \to v \to r$ means consecutive integrations. Because the initial state is from rest at origin, we don't have to worry about constants of integration.

Chapter Mix

Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

Q43 jee_main_2026_21_jan_evening Circular Motion

A large drum having radius R is spinning around its axis with angular velocity

ω

, as shown in figure. The minimum value of

ω

so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M is

μ

, is :

Spinning drum with mass M for Q43 - JEE Main 2026 Evening — Large drum of radius R spinning about its vertical axis with mass M stuck to the inner wall.

A. $\sqrt((μ g)/(R))$ $\sqrt{\frac{\mu g}{R}}$
B. $\sqrt((2g)/(μ R))$ $\sqrt{\frac{2g}{\mu R}}$
C. $\sqrt((g)/(2μ R))$ $\sqrt{\frac{g}{2\mu R}}$
D. $\sqrt((g)/(μ R))$ $\sqrt{\frac{g}{\mu R}}$

Solution

Related Formula

$N = Mω² R$ $N = M\omega^2 R$

fmax = μ N

$f = Mg$ $f = Mg$

Core Logic

Free body diagram for spinning drum Q43 - JEE Main 2026 Evening — Large drum of radius R spinning about its vertical axis with mass M stuck to the inner wall.

For the mass $M$ $M$ to remain stuck to the inner wall and not slide down, the upward frictional force must balance the downward gravitational force. $f = Mg$ $f = Mg$ The normal force $N$ $N$ providing the friction is generated by the centrifugal effect (or provides the centripetal acceleration): $N = Mω² R$ $N = M\omega^2 R$ The condition for no sliding is that required friction cannot exceed the maximum static friction: $Mg \leq μ N$ $Mg \le \mu N$

Step 1: Final Conclusion

Substitute $N$ $N$ into the inequality:

Mg \leq μ (Mω² R)

g \leq μ ω² R

ω² \geq (g)/(μ R)

For minimum angular velocity:

ωmin = \sqrt((g)/(μ R))

Pattern Recognition

In a "rotor" ride setup, gravity is opposed by friction, and the normal force is solely centripetal. Equating $μ Fc$ $\mu F_c$ to $mg$ $mg$ always yields the minimum spin limit.

Chapter Mix

Class 11 Physics: Laws of Motion

Q28 jee_main_2026_23_january_evening Motion on Inclined Plane

A block is sliding down on an inclined plane of slope

θ

and at an instant

t = 0

this block is given an upward momentum so that it starts moving up on the inclined surface with velocity

u

. The distance (S) travelled by the block before its velocity become zero, is ____. (g = gravitational acceleration)

A. $(u²)/(4g θ)$ $\frac{u^2}{4g\sin\theta}$
B. $(2u²)/(g θ)$ $\frac{2u^2}{g\cos\theta}$
C. $u²\sqrt(2)g θ$ $\frac{u^2}{\sqrt{2}g\cos\theta}$
D. $(u²)/(2g θ)$ $\frac{u^2}{2g\sin\theta}$

Solution

Related Formula

$v² = u² + 2as$ $v^2 = u^2 + 2as$

Core Logic

Motion on Inclined Plane diagram for Q28 - JEE Main 2026 Evening

Since the surface is frictionless (implied as no friction coefficient is given), the only force along the incline opposing the upward motion is the component of gravity, $mg θ$ $mg \sin \theta$.

Acceleration $a = -g θ$ $a = -g \sin \theta$.

Step 1: Apply Kinematic Equation

Using the third equation of motion: $V² = U² + 2as$ $V^2 = U^2 + 2as$

Set final velocity $V = 0$ $V = 0$:

0 = u² - 2(g θ) S

S = (u²)/(2g θ)

Pattern Recognition

For upward motion on a smooth incline, deceleration is purely $g θ$ $g \sin \theta$. The stopping distance is always $u² / 2a$ $u^2 / 2a$.

Chapter Mix

Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

Q36 jee_main_2026_24_january_morning Spring Force

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1:3, then the force constant of smaller piece is ____ N/m

A. $15$ $15$
B. $20$ $20$
C. $60$ $60$
D. $45$ $45$

Solution

Related Formula

K = constant

K \propto (1)/( )

Core Logic

The spring of total length $\ell$ is cut into two parts with a length ratio of 1:3. The lengths of the two pieces are: $₁ = (1)/(4)$ $\ell_1 = \frac{1}{4} \ell$ (smaller piece) $₂ = (3)/(4)$ $\ell_2 = \frac{3}{4} \ell$ (larger piece)

Step 1: Calculate Spring Constant of Smaller Piece

Since $K = K' '$ $K \ell = K' \ell'$, for the smaller piece:

K = K' (( )/(4))

$K' = 4K$ $K' = 4K$ Given $K = 15 N/m$ $K = 15 \text{ N/m}$:

K' = 4 \times 15 = 60 N/m

Pattern Recognition

Spring constant is inversely proportional to spring length. Cutting a spring to $1/n$ $1/n$ of its length increases its stiffness by a factor of $n$ $n$.

Chapter Mix

Class 11 Physics: Laws of Motion

Q49 jee_main_2026_24_january_morning Friction

In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force

F

required to slide the block C with constant speed is ____ N. (Used

g = 10 m/s²

)

Stacked blocks with friction and string — Three stacked blocks A, B, C with an external pulling force F applied to C.

Numerical Answer. Answer: 210 to 210

Solution

Related Formula

$fk = μk N$ $f_k = \mu_k N$

Fₙₑₜ = 0 (for constant speed)

Core Logic

Free body diagram for stacked blocks — Three stacked blocks A, B, C with an external pulling force F applied to C.

Since C moves with constant speed, $a=0$ $a=0$, meaning net force is zero. For block A ( $4 kg$ $4\text{ kg}$) over B, sliding friction $fA = μ (mA)g = 0.5 \times 4 \times 10 = 20 N$ . For block B ( $6 kg$ $6\text{ kg}$) over C, the normal force includes block A. $NB = (mA + mB)g = 100 N$ $N_B = (m_A + m_B)g = 100\text{ N}$. Friction on B from C is $fB = 0.5 \times 100 = 50 N$ $f_B = 0.5 \times 100 = 50\text{ N}$. For block C ( $8 kg$ $8\text{ kg}$) over ground, normal force is total weight $NC = (mA + mB + mC)g = 180 N$ . Friction on C from ground is $fC = 0.5 \times 180 = 90 N$ $f_C = 0.5 \times 180 = 90\text{ N}$.

Step 1: Force Balance Equations

For block B to move relative to A and C (assuming standard string configuration where pulling C left causes B to be dragged right relative to C via pulley): The tension $T$ $T$ on B balances friction from A and C:

T = fA + fB = 20 + 50 = 70 N

For block C moving left, the force $F$ $F$ must overcome friction from ground, friction from B, and the tension $T$ $T$ attached to it (depending on pulley setup. If pulley is on wall, string pulls C right): Assuming standard setup where string connects B and C around wall pulley:

F = fC + fB + T

F = 90 + 50 + 70 = 210 N

Pattern Recognition

In multi-block pulley systems, pulling the bottom block accumulates all nested friction forces. The bottom block fights the ground, the block above it, AND the tension of the string anchoring the block above it.

Chapter Mix

Class 11 Physics: Laws of Motion

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Friction

Step 2: Force Equations

Step 3: Difference calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Laws of Motion Previous-Year Questions

Solution

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Step 2: Calculating Position

Pattern Recognition

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Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Kinematic Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Spring Constant of Smaller Piece

Pattern Recognition

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Step 1: Force Balance Equations

Pattern Recognition

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