A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving…

Year	2026	2025	2024	Total
Questions	9	10	15	34

Q39 jee_main_2024_29_jan_morning Friction and Work Done

A block of mass

100 ~kg

slides over a distance of

10 ~m

on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction (in J) is:

A. 4200
B. 3900
C. 4000
D. 4500

Solution

Related Formula

Kinetic frictional force ( $f$ $f$) on a flat surface:

f = μ N = μ m g

Work done against friction ( $Wagainst$ $W_{\text{against}}$):

Wagainst = f \cdot s = μ m g s

Core Logic

Given values:

m = 100 ~kg, s = 10 ~m, μ = 0.4

Take $g = 10 ~m/s²$ $g = 10 \mathrm{~m/s^2}$.

Step 1: Calculate Friction Force

The normal force on a horizontal surface is:

N = mg = 100 \times 10 = 1000 ~N

The force of friction is:

f = μ N = 0.4 \times 1000 = 400 ~N

Step 2: Calculate Work Done

The work done against friction is:

Wagainst = f \cdot s = 400 ~N \times 10 ~m = 4000 ~J

Therefore, the work done is $4000 ~J$ $4000 \mathrm{~J}$.

Pattern Recognition

Work done by friction is negative ( $-4000 ~J$ $-4000 \mathrm{~J}$) because the frictional force opposes displacement. Work done against friction is positive ( $+4000 ~J$ $+4000 \mathrm{~J}$) because it represents the external energy that must be spent to sustain slide.

Chapter Mix

Class 11 Physics: Laws of Motion

Q jee_main_2024_30_january_evening Connected Bodies and Tension

Three blocks

A

,

B

and

C

are pulled on a horizontal smooth surface by a force of

80N

as shown in figure

Connected Bodies and Tension diagram for Q47 - JEE Main 2024 Evening — Three masses (5 kg, 3 kg, 2 kg) connected by strings with tensions T1 and T2, being pulled by a common force F = 80N.

The tensions

T₁

and

T₂

in the string are respectively:

A. $40 N, 64 N$ $40 \mathrm{N}, 64 \mathrm{N}$
B. $60 N, 80 N$ $60 \mathrm{N}, 80 \mathrm{N}$
C. $88 N, 96 N$ $88 \mathrm{N}, 96 \mathrm{N}$
D. $80 N, 100 N$ $80 \mathrm{N}, 100 \mathrm{N}$

Solution

Related Formula

a = Fₙₑₜmtotal

T = mbehind \times a

Core Logic

Since the surface is smooth, the entire system of three blocks moves with a common acceleration $a$ $a$.

aA = aB = aC = FmA + mB + mC

Step 1: Calculate Common Acceleration

a = (80)/(5 + 3 + 2) = (80)/(10) = 8 ~m/s²

Step 2: Calculate Tensions

Connected Bodies and Tension — Three masses (5 kg, 3 kg, 2 kg) connected by strings with tensions T1 and T2, being pulled by a common force F = 80N.

$T₁$ $\mathrm{T}_1$ pulls only block $A$ $\mathrm{A}$ ( $5 ~kg$ $5 \mathrm{~kg}$):

T₁ = mA \times a = 5 \times 8 = 40 ~N

$T₂$ $\mathrm{T}_2$ pulls both blocks $A$ $\mathrm{A}$ and $B$ $\mathrm{B}$ ( $5 ~kg + 3 ~kg = 8 ~kg$ ): From free body diagram of block B:

T₂ - T₁ = mB \times a

T₂ - 40 = 3 \times 8 = 24 T₂ = 64 ~N

Pattern Recognition

Tension at any point in a train of connected accelerating bodies (without friction) is directly proportional to the total mass being pulled behind that point.

Chapter Mix

Class 11 Physics: Laws of Motion

Q32 jee_main_2024_30_january_evening Work Done by Friction on Incline

A block of mass

1 ~kg

is pushed up a surface inclined to horizontal at an angle of

60°

by a force of

10 ~N

parallel to the inclined surface as shown in figure. When the block is pushed up by

10 ~m

along inclined surface, the work done against frictional force is:

[g = 10 ~m / s²]

Work Done by Friction on Incline diagram for Q32 - JEE Main 2024 Evening — A block of mass M on an incline at 60 degrees, pulled by 10 N force, with coefficient of static friction 0.1.

A. $5\sqrt(3) ~J$ $5\sqrt{3} \mathrm{~J}$
B. $5 ~J$ $5 \mathrm{~J}$
C. $5 \times 10³ ~J$ $5 \times 10^{3} \mathrm{~J}$
D. $10 ~J$ $10 \mathrm{~J}$

Solution

Related Formula

Wf = fk \cdot d

$fk = μ N$ $f_k = \mu N$

N = mg θ

Core Logic

The work done against the frictional force is the product of the kinetic friction force and the displacement along the plane. The normal force $N$ $N$ on the block is given by $N = mg θ$ $N = mg \cos\theta$, where $θ = 60°$ $\theta = 60^{\circ}$.

Step 1: Calculate Normal and Frictional Force

Given: $μ = 0.1$ $\mu = 0.1$ $m = 1 ~kg$ $m = 1 \mathrm{~kg}$ $θ = 60°$ $\theta = 60^{\circ}$ $d = 10 ~m$ $d = 10 \mathrm{~m}$

Normal force $N = 1 \times 10 \times (60°) = 10 \times (1)/(2) = 5 ~N$ . Frictional force $fk = μ N = 0.1 \times 5 = 0.5 ~N$ .

Step 2: Work Done Against Friction

Work done against frictional force $= fk \times d$ $= f_k \times d$

W = 0.5 ~N \times 10 ~m = 5 ~J

Pattern Recognition

Whenever asked for "work done against friction", simply compute $μ mg θ \times d$ $\mu mg \cos\theta \times d$. The applied force ( $10 ~N$ $10 \mathrm{~N}$) is irrelevant to the friction calculation itself since it is parallel to the plane.

Chapter Mix

Class 11 Physics: Laws of Motion Class 11 Physics: Work, Energy and Power

Q39 jee_main_2024_30_january_evening Equilibrium on a Rough Parabolic Curve

A block of mass

m

is placed on a surface having vertical cross section given by

y = x² / 4

. If coefficient of friction is

0.5

, the maximum height above the ground at which block can be placed without slipping is:

A. $1 / 4 ~m$ $1 / 4 \mathrm{~m}$
B. $1 / 2 ~m$ $1 / 2 \mathrm{~m}$
C. $1 / 6 ~m$ $1 / 6 \mathrm{~m}$
D. $1 / 3 ~m$ $1 / 3 \mathrm{~m}$

Solution

Related Formula

θ = μ

dydx = θ

Core Logic

For a block to remain stationary on a rough surface without slipping, the maximum slope of the surface it can rest on is determined by the angle of repose.

θ \leq μ

The slope of the given parabolic curve at any point $(x,y)$ $(x,y)$ is $dydx$ $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Step 1: Evaluate the Slope

Given curve equation:

y = (x²)/(4)

Differentiating with respect to $x$ $x$:

dydx = (2x)/(4) = (x)/(2)

Step 2: Find Maximum Valid Position

At the critical point of slipping, slope equals $μ$ $\mu$:

(x)/(2) = μ

Given $μ = 0.5 = (1)/(2)$ $\mu = 0.5 = \frac{1}{2}$, we have:

(x)/(2) = (1)/(2) x = 1

Step 3: Calculate Maximum Height

Substitute $x = 1$ $x = 1$ back into the curve equation to find the maximum height $y$ $y$:

y = ((1)²)/(4) = (1)/(4) ~m

Pattern Recognition

When asked for maximum height on a curve $y=f(x)$ $y=f(x)$ without slipping, set $dydx = μ$ $\frac{\mathrm{d}y}{\mathrm{d}x} = \mu$, solve for $x$ $x$, and plug it back into the original equation to find $y$ $y$.

Chapter Mix

Class 11 Physics: Laws of Motion

Q32 jee_main_2024_30_jan_morning Constraint Motion and Pulleys

All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass

2 ~kg

is:

Constraint Motion and Pulleys diagram for Q32 - JEE Main 2024 Morning — Illustration of a 2kg block on a 30 degree incline attached via a pulley system to a 4kg hanging block.

A. $g$ $g$
B. $(g)/(3)$ $\frac{g}{3}$
C. $(g)/(2)$ $\frac{g}{2}$
D. $(g)/(4)$ $\frac{g}{4}$

Solution

Related Formula

Σ F = ma

a₁ = Constraint relation\times a₂

Core Logic

Solution schematic showing free body diagrams and tensions. — Illustration of a 2kg block on a 30 degree incline attached via a pulley system to a 4kg hanging block.

Let the tension in the string attached to the

2 ~kg

block be

T

. By tracing the string around the movable pulley, the tension supporting the

4 ~kg

mass becomes

2T

.

From the principle of virtual work (or string constraints), if the $4 ~kg$ $4 \mathrm{~kg}$ block moves down with an acceleration $a$ $a$, the string shortens by $2x$ $2x$ on the incline side, meaning the $2 ~kg$ $2 \mathrm{~kg}$ block moves up the incline with an acceleration of $2a$ $2a$.

Step 1: Write Equations of Motion

For the $4 ~kg$ $4 \mathrm{~kg}$ block (moving downwards): $4g - 2T = 4a$ $4g - 2T = 4a$

40 - 2T = 4a (i)

For the $2 ~kg$ $2 \mathrm{~kg}$ block (moving up the incline):

T - 2g (30^°) = 2(2a)

T - 2(10)((1)/(2)) = 4a

T - 10 = 4a (ii)

Step 2: Solve for Acceleration

From equation (ii), we have $4a = T - 10$ $4a = T - 10$. Substitute this directly into equation (i) or add the appropriately multiplied equations:

40 - 2T = T - 10

3T = 50 \Rightarrow T = (50)/(3) ~N

Substitute $T$ $T$ back into (ii):

(50)/(3) - 10 = 4a

4a = (20)/(3) \Rightarrow a = (5)/(3) ~m/s²

We need the acceleration of the $2 ~kg$ $2 \mathrm{~kg}$ block, which is $2a$ $2a$:

2a = 2 \times ((5)/(3)) = (10)/(3) ~m/s²

Since $g = 10 ~m/s²$ $g = 10 \mathrm{~m/s^2}$, this acceleration is exactly $(g)/(3)$ $\frac{g}{3}$.

Pattern Recognition

Movable pulleys double the force but halve the displacement/acceleration on the supported side. Tension on the movable pulley side is twice the tension on the single string side. Remember to explicitly solve for the requested block's specific acceleration, not just 'a'.

Chapter Mix

Class 11 Physics: Laws of Motion

Solution & Explanation

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

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Calculate Friction Force

Step 2: Calculate Work Done

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Common Acceleration

Step 2: Calculate Tensions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Normal and Frictional Force

Step 2: Work Done Against Friction

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate the Slope

Step 2: Find Maximum Valid Position

Step 3: Calculate Maximum Height

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Write Equations of Motion

Step 2: Solve for Acceleration

Pattern Recognition

Chapter Mix

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