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

Q40 jee_main_2026_24_january_evening Equilibrium of Forces

A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is

30°

. Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ____.

A. $\sqrt(3)2 m g$ $\frac{\sqrt{3}}{2} m g$
B. $(1)/(2) m g$ $\frac{1}{2} m g$
C. $m g$ $m g$
D. $\sqrt(3) m g$ $\sqrt{3} m g$

Solution

Related Formula

Σ Fy = 0 T θ = (mg)/(2)

Σ Fₓ = 0 T θ = T₀

Core Logic

Equilibrium of Forces diagram for Q40 - JEE Main 2026 Evening

Draw the Free Body Diagram (F.B.D) of half of the rope. The forces acting on half the rope (mass $m/2$ $m/2$) are:

Weight $(mg)/(2)$ $\frac{mg}{2}$ acting downwards.
Tension $T$ $T$ at the support point acting at $30°$ $30^{\circ}$ to the horizontal.
Horizontal tension $T₀$ $T_0$ at the lowest point.

Step 1: Equilibrium Equations

For vertical equilibrium:

T 30° = (m)/(2) g

For horizontal equilibrium:

T 30° = T₀

Step 2: Solve for T_0

Dividing the vertical equation by the horizontal equation:

30° = (mg / 2)/(T₀)

T₀ = mg2 30°

T₀ = mg2 (1/\sqrt(3)) = \sqrt(3)2 mg

Pattern Recognition

When dealing with symmetrical hanging chains, always cut the chain at the lowest point. The tension at the lowest point is purely horizontal and is given by $T₀ = (W/2) / θ$ $T_0 = (W/2) / \tan\theta$, where $θ$ $\theta$ is the angle at the supports.

Chapter Mix

Class 11 Physics: Laws of Motion

Q40 jee_main_2026_28_january_morning Friction on an Inclined Plane

A block of mass 5 kg is moving on an inclined plane which makes an angle of

30°

with the horizontal. Friction coefficient between the block and inclined plane surface is

\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

Related Formula

fk = μk N = μk mg θ

Fdown = mg θ

$Net force = 0 for constant velocity$

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°)

Step 2: Value Substitution

F = μ mg (30°) - mg (30°)

F = ( \sqrt(3)2 ) (5 \times 10) ( \sqrt(3)2 ) - (5 \times 10) ( (1)/(2) )

Step 3: Final Calculation

F = 50 \times (3)/(4) - 50 \times (1)/(2)

F = 37.5 - 25 = 12.5 ~N

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 $\sqrt(3)2 = 0.866$ $\frac{\sqrt{3}}{2} = 0.866$ and $30^° = 1\sqrt(3) \approx 0.577$ ), friction is stronger than gravity's pull. You must actively pull it down to keep it moving.

Chapter Mix

Class 11 Physics: Laws of Motion

Q45 jee_main_2026_28_january_morning Motion with Drag Force

A particle of mass m falls from rest through a resistive medium having resistive force,

F = -kv

, where v is the velocity of the particle and k is a constant. Which of the following graphs represents velocity (v) versus time(t)?

A. Graph 1
B. Graph 2
C. Graph 3
D. Graph 4

Solution

Related Formula

Fₙₑₜ = ma

mg - kv = m(dv)/(dt)

Core Logic

Integrate the differential equation of motion to find how velocity depends on time. Initially, acceleration is $g$ $g$. As velocity increases, the resistive force $kv$ $kv$ increases, reducing acceleration until it reaches zero (terminal velocity).

Step 1: Setting up Integration

m \cdot (dv)/(dt) = mg - kv

\int₀v (dv)/(mg - kv) = \int₀t (dt)/(m)

Step 2: Performing Integration

[ -(1)/(k) ln(mg - kv) ]₀^v = (t)/(m)

-(1)/(k) ln((mg - kv)/(mg)) = (t)/(m)

ln(1 - (kv)/(mg)) = -(kt)/(m)

Step 3: Final Velocity Equation

1 - (kv)/(mg) = e-kt/m

v = (mg)/(k) (1 - e-kt/m)

This shows an exponential increase that asymptotically approaches a terminal velocity $vterminal = mg/k$ $v_{\text{terminal}} = mg/k$. Option (2) represents this curve.

Pattern Recognition

The equation $1 - e-t$ $1 - e^{-t}$ always produces a curve starting at origin and flattening out horizontally (asymptotic approach to a limit).

Chapter Mix

Class 11 Physics: Laws of Motion

Q40 jee_main_2026_28_january_evening Pseudo Force and Inclined Plane

A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration

a₀

. The angle between the inclined plane and ground is

θ

and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ____.

Pseudo Force and Inclined Plane diagram for Q40 - JEE Main 2026 Evening — A mass m on an inclined plane of base L, accelerating to the left with a0.

A. $\sqrt((2L)/(g 2θ - a₀(1 + 2θ)))$
B. $\sqrt((4L)/(g 2θ - a₀(1 + 2θ)))$
C. $\sqrt((4L)/(g ²θ - a₀ θ θ))$
D. $\sqrt((2L)/(g θ - a₀ θ))$

Solution

Related Formula

s = ut + (1)/(2)at²

Fpseudo = m a₀

Core Logic

Solution for Pseudo Force and Inclined Plane — A mass m on an inclined plane of base L, accelerating to the left with a0.

We solve the problem from the non-inertial frame of reference of the inclined plane. A pseudo force

m a₀

acts on the block towards the right. Forces acting parallel to the incline (downward positive):

Component of gravity: $mg θ$ $mg \sin \theta$
Component of pseudo force (pointing up the incline): $-ma₀ θ$ $-ma_0 \cos \theta$

Step 1: Calculate Effective Acceleration

Net force down the incline:

Fₙₑₜ = mg θ - m a₀ θ

Thus, the acceleration relative to the incline is:

a = g θ - a₀ θ

Step 2: Distance Travelled

The base length of the incline is $L$ $L$. Thus, the total length of the inclined surface is:

S = (L)/( θ)

Step 3: Calculate Time

Using the second equation of motion ( $u=0$ $u=0$):

S = (1)/(2) a t²

(L)/( θ) = (1)/(2) (g θ - a₀ θ) t²

t² = (2L)/( θ (g θ - a₀ θ))

t = \sqrt( (2L)/(g θ θ - a₀ ² θ) )

Step 4: Trigonometric Identity Substitution

Multiply numerator and denominator by 2 inside the square root:

t = \sqrt( (4L)/(2g θ θ - 2a₀ ² θ) )

Using $2θ = 2 θ θ$ $\sin 2\theta = 2\sin\theta\cos\theta$ and $2 ²θ = 1 + 2θ$ $2\cos^2\theta = 1 + \cos 2\theta$:

t = \sqrt( (4L)/(g 2θ - a₀(1 + 2θ)) )

Pattern Recognition

Whenever an inclined plane accelerates horizontally, resolve the pseudo force $ma₀$ $ma_0$ parallel and perpendicular to the incline. Multiply by 2 inside the root to simplify into double-angle formats if options demand it.

Chapter Mix

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

Q16 jee_main_2025_02_april_evening Equilibrium of Forces

A body of mass

1kg

is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions

T₁

and

T₂

, respectively, are (in N):

Suspended mass equilibrium with two angled strings — The diagram shows a suspended mass of 1 kg held by two strings making angles of 60 and 30 degrees with the horizontal.

A. $5, 5\sqrt(3)$ $5, 5\sqrt{3}$
B. $5\sqrt(3), 5$ $5\sqrt{3}, 5$
C. $5\sqrt(3), 5\sqrt(3)$ $5\sqrt{3}, 5\sqrt{3}$
D. $5, 5$ $5, 5$

Solution

Related Formula

For a system in static equilibrium:

Σ Fₓ = 0 and Σ Fy = 0

Core Logic

Let's resolve the tension forces $T₁$ $\vec{T}_1$ and $T₂$ $\vec{T}_2$ into horizontal and vertical components:

$T₁$ $T_1$ makes $60^°$ $60^\circ$ with the horizontal.
$T₂$ $T_2$ makes $30^°$ $30^\circ$ with the horizontal.
Downward gravitational force: $W = m g = 1 \times 10 = 10 N$ .
Horizontal Equilibrium ( $Σ Fₓ = 0$ $\sum F_x = 0$):

T₁ 60^° = T₂ 30^°

T₁ \cdot (1)/(2) = T₂ \cdot \sqrt(3)2 T₁ = T₂ \sqrt(3)

Vertical Equilibrium ( $Σ Fy = 0$ $\sum F_y = 0$):

T₁ 60^° + T₂ 30^° = m g = 10

T₁ \cdot \sqrt(3)2 + T₂ \cdot (1)/(2) = 10

Step 1: Solve for Tensions

Substitute $T₁ = T₂ \sqrt(3)$ $T_1 = T_2 \sqrt{3}$ into the vertical equilibrium equation:

(T₂ \sqrt(3)) \sqrt(3)2 + (T₂)/(2) = 10

(3 T₂)/(2) + (T₂)/(2) = 10 2 T₂ = 10 T₂ = 5 N

Substitute $T₂$ $T_2$ back to obtain $T₁$ $T_1$:

T₁ = 5 \sqrt(3) N

Thus, the tension magnitudes are $T₁ = 5\sqrt(3) N$ $T_1 = 5\sqrt{3} \ \mathrm{N}$ and $T₂ = 5 N$ $T_2 = 5 \ \mathrm{N}$.

Pattern Recognition

Sees: Suspending particle static equilibrium with asymmetric strings. Trap: Associating components with incorrect trigonometry axes or swapping $T₁$ $T_1$ and $T₂$ $T_2$ in options. Shortcut: Since the incline of $T₁$ $T_1$ ( $60^°$ $60^\circ$) is steeper than that of $T₂$ $T_2$ ( $30^°$ $30^\circ$), $T₁$ $T_1$ must carry a larger portion of the load, meaning $T₁ > T₂$ $T_1 > T_2$. From the choices, only (2) satisfies this hierarchy.

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 — Page 2

Solution

Related Formula

Core Logic

Step 1: Equilibrium Equations

Step 2: Solve for T_0

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Force Balance Equation

Step 2: Value Substitution

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up Integration

Step 2: Performing Integration

Step 3: Final Velocity Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Effective Acceleration

Step 2: Distance Travelled

Step 3: Calculate Time

Step 4: Trigonometric Identity Substitution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for Tensions

Pattern Recognition

Chapter Mix

More Laws of Motion Questions — jee_main_2024_31_jan_evening

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