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 mathrm~kg$2 \mathrm{~kg}$ is:
Illustration of a 2kg block on a 30 degree incline attached via a pulley system to a 4kg hanging block.
A.g$g$
B.fracg3$\frac{g}{3}$
C.fracg2$\frac{g}{2}$
D.fracg4$\frac{g}{4}$
Solution & Explanation
### Related Formula
sum F = ma$\sum F = ma$a_1 = textConstraint relationtimes a_2$a_1 = \text{Constraint relation}\times a_2$
### Core Logic
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 mathrm~kg$2 \mathrm{~kg}$ block be T$T$. By tracing the string around the movable pulley, the tension supporting the 4 mathrm~kg$4 \mathrm{~kg}$ mass becomes 2T$2T$.
From the principle of virtual work (or string constraints), if the 4 mathrm~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 mathrm~kg$2 \mathrm{~kg}$ block moves up the incline with an acceleration of 2a$2a$.
### Step 1: Write Equations of Motion
For the 4 mathrm~kg$4 \mathrm{~kg}$ block (moving downwards):
4g - 2T = 4a$4g - 2T = 4a$40 - 2T = 4a quad dots (i)$40 - 2T = 4a \quad \dots (i)$
For the 2 mathrm~kg$2 \mathrm{~kg}$ block (moving up the incline):
T - 2g sin(30^circ) = 2(2a)$T - 2g \sin(30^\circ) = 2(2a)$T - 2(10)left(frac12right) = 4a$T - 2(10)\left(\frac{1}{2}\right) = 4a$T - 10 = 4a quad dots (ii)$T - 10 = 4a \quad \dots (ii)$
### 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$40 - 2T = T - 10$3T = 50 Rightarrow T = frac503 mathrm~N$3T = 50 \Rightarrow T = \frac{50}{3} \mathrm{~N}$
Substitute T$T$ back into (ii):
frac503 - 10 = 4a$\frac{50}{3} - 10 = 4a$4a = frac203 Rightarrow a = frac53 mathrm~m/s^2$4a = \frac{20}{3} \Rightarrow a = \frac{5}{3} \mathrm{~m/s^2}$
We need the acceleration of the 2 mathrm~kg$2 \mathrm{~kg}$ block, which is 2a$2a$:
2a = 2 times left(frac53right) = frac103 mathrm~m/s^2$2a = 2 \times \left(\frac{5}{3}\right) = \frac{10}{3} \mathrm{~m/s^2}$
Since g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$, this acceleration is exactly fracg3$\frac{g}{3}$.
### 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'.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Keywords:#Constraint Motion and Pulleys#JEE Main 2024 Morning Q32#Laws of Motion JEE Main 2024#Pulley acceleration#Pulley system#Inclined plane#Mass blocks#String tension
More Laws of Motion Previous-Year Questions — Page 4
Q45jee_main_2024_27_jan_morningConservation of Linear Momentum
A body of mass 1000text kg$1000\text{ kg}$ is moving horizontally with a velocity 6text m/s$6\text{ m/s}$. If 200text 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
m_1 v_1 = m_2 v_2$m_1 v_1 = m_2 v_2$
### Core Logic
Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved.
Initial mass m_1 = 1000text kg$m_1 = 1000\text{ kg}$, initial velocity v_1 = 6text m/s$v_1 = 6\text{ m/s}$.
Final mass m_2 = 1000 + 200 = 1200text kg$m_2 = 1000 + 200 = 1200\text{ kg}$.
### Step 1: Balance conservation equation
1000 times 6 = 1200 times v_2$1000 \times 6 = 1200 \times v_2$6000 = 1200 v_2 implies v_2 = frac60001200 = 5text m/s$6000 = 1200 v_2 \implies v_2 = \frac{6000}{1200} = 5\text{ m/s}$
### Pattern Recognition
Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q39jee_main_2024_29_jan_morningFriction and Work Done
A block of mass 100 mathrm~kg$100 \mathrm{~kg}$ slides over a distance of 10 mathrm~m$10 \mathrm{~m}$ 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 = mu N = mu m g$f = \mu N = \mu m g$Work done against friction (W_textagainst$W_{\text{against}}$):
W_textagainst = f cdot s = mu m g s$W_{\text{against}} = f \cdot s = \mu m g s$
### Core Logic
Given values:
m = 100 mathrm~kg, quad s = 10 mathrm~m, quad mu = 0.4$m = 100 \mathrm{~kg}, \quad s = 10 \mathrm{~m}, \quad \mu = 0.4$
Take g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$.
### Step 1: Calculate Friction Force
The normal force on a horizontal surface is:
N = mg = 100 times 10 = 1000 mathrm~N$N = mg = 100 \times 10 = 1000 \mathrm{~N}$
The force of friction is:
f = mu N = 0.4 times 1000 = 400 mathrm~N$f = \mu N = 0.4 \times 1000 = 400 \mathrm{~N}$
### Step 2: Calculate Work Done
The work done against friction is:
W_textagainst = f cdot s = 400 mathrm~N times 10 mathrm~m = 4000 mathrm~J$W_{\text{against}} = f \cdot s = 400 \mathrm{~N} \times 10 \mathrm{~m} = 4000 \mathrm{~J}$
Therefore, the work done is 4000 mathrm~J$4000 \mathrm{~J}$.
### Pattern Recognition
Work done *by* friction is negative (-4000 mathrm~J$-4000 \mathrm{~J}$) because the frictional force opposes displacement. Work done *against* friction is positive (+4000 mathrm~J$+4000 \mathrm{~J}$) because it represents the external energy that must be spent to sustain slide.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q32jee_main_2024_30_january_eveningWork Done by Friction on Incline
A block of mass 1 mathrm~kg$1 \mathrm{~kg}$ is pushed up a surface inclined to horizontal at an angle of 60^circ$60^{\circ}$ by a force of 10 mathrm~N$10 \mathrm{~N}$ parallel to the inclined surface as shown in figure. When the block is pushed up by 10 mathrm~m$10 \mathrm{~m}$ along inclined surface, the work done against frictional force is: left[mathrmg = 10 mathrm~m / mathrms^2right]$\left[\mathrm{g} = 10 \mathrm{~m} / \mathrm{s}^{2}\right]$A block of mass M on an incline at 60 degrees, pulled by 10 N force, with coefficient of static friction 0.1.
A.5sqrt3 mathrm~J$5\sqrt{3} \mathrm{~J}$
B.5 mathrm~J$5 \mathrm{~J}$
C.5 times 10^3 mathrm~J$5 \times 10^{3} \mathrm{~J}$
D.10 mathrm~J$10 \mathrm{~J}$
Solution
### Related Formula
W_f = f_k cdot d$W_f = f_k \cdot d$f_k = mu N$f_k = \mu N$N = mg costheta$N = mg \cos\theta$
### 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 costheta$N = mg \cos\theta$, where theta = 60^circ$\theta = 60^{\circ}$.
### Step 1: Calculate Normal and Frictional Force
Given:
mu = 0.1$\mu = 0.1$m = 1 mathrm~kg$m = 1 \mathrm{~kg}$theta = 60^circ$\theta = 60^{\circ}$d = 10 mathrm~m$d = 10 \mathrm{~m}$
Normal force N = 1 times 10 times cos(60^circ) = 10 times frac12 = 5 mathrm~N$N = 1 \times 10 \times \cos(60^{\circ}) = 10 \times \frac{1}{2} = 5 \mathrm{~N}$.
Frictional force f_k = mu N = 0.1 times 5 = 0.5 mathrm~N$f_k = \mu N = 0.1 \times 5 = 0.5 \mathrm{~N}$.
### Step 2: Work Done Against Friction
Work done against frictional force= f_k times d$= f_k \times d$W = 0.5 mathrm~N times 10 mathrm~m = 5 mathrm~J$W = 0.5 \mathrm{~N} \times 10 \mathrm{~m} = 5 \mathrm{~J}$
### Pattern Recognition
Whenever asked for "work done against friction", simply compute mu mg costheta times d$\mu mg \cos\theta \times d$. The applied force (10 mathrm~N$10 \mathrm{~N}$) is irrelevant to the friction calculation itself since it is parallel to the plane.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Work, Energy and Power
Q39jee_main_2024_30_january_eveningEquilibrium on a Rough Parabolic Curve
A block of mass m$m$ is placed on a surface having vertical cross section given by y = x^2 / 4$y = x^2 / 4$. If coefficient of friction is 0.5$0.5$, the maximum height above the ground at which block can be placed without slipping is:
A.1 / 4 mathrm~m$1 / 4 \mathrm{~m}$
B.1 / 2 mathrm~m$1 / 2 \mathrm{~m}$
C.1 / 6 mathrm~m$1 / 6 \mathrm{~m}$
D.1 / 3 mathrm~m$1 / 3 \mathrm{~m}$
Solution
### Related Formula
tan theta = mu$\tan \theta = \mu$fracmathrmdymathrmdx = tan theta$\frac{\mathrm{d}y}{\mathrm{d}x} = \tan \theta$
### 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.
tan theta le mu$\tan \theta \le \mu$
The slope of the given parabolic curve at any point (x,y)$(x,y)$ is fracmathrmdymathrmdx$\frac{\mathrm{d}y}{\mathrm{d}x}$.
### Step 1: Evaluate the Slope
Given curve equation:
y = fracx^24$y = \frac{x^2}{4}$
Differentiating with respect to x$x$:
fracmathrmdymathrmdx = frac2x4 = fracx2$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x}{4} = \frac{x}{2}$
### Step 2: Find Maximum Valid Position
At the critical point of slipping, slope equals mu$\mu$:
fracx2 = mu$\frac{x}{2} = \mu$
Given mu = 0.5 = frac12$\mu = 0.5 = \frac{1}{2}$, we have:
fracx2 = frac12 implies x = 1$\frac{x}{2} = \frac{1}{2} \implies x = 1$
### Step 3: Calculate Maximum Height
Substitute x = 1$x = 1$ back into the curve equation to find the maximum height y$y$:
y = frac(1)^24 = frac14 mathrm~m$y = \frac{(1)^2}{4} = \frac{1}{4} \mathrm{~m}$
### Pattern Recognition
When asked for maximum height on a curve y=f(x)$y=f(x)$ without slipping, set fracmathrmdymathrmdx = mu$\frac{\mathrm{d}y}{\mathrm{d}x} = \mu$, solve for x$x$, and plug it back into the original equation to find y$y$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q47jee_main_2024_30_january_eveningConnected Bodies and Tension
Three blocks mathrmA$\mathrm{A}$, mathrmB$\mathrm{B}$ and mathrmC$\mathrm{C}$ are pulled on a horizontal smooth surface by a force of 80mathrmN$80\mathrm{N}$ as shown in figure
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 mathrmT_1$\mathrm{T}_{1}$ and mathrmT_2$\mathrm{T}_{2}$ in the string are respectively:
### Related Formula
a = fracF_textnetm_texttotal$a = \frac{F_{\text{net}}}{m_{\text{total}}}$T = m_textbehind times a$T = m_{\text{behind}} \times a$
### Core Logic
Since the surface is smooth, the entire system of three blocks moves with a common acceleration a$a$.
a_mathrmA = a_mathrmB = a_mathrmC = fracFm_mathrmA + m_mathrmB + m_mathrmC$a_{\mathrm{A}} = a_{\mathrm{B}} = a_{\mathrm{C}} = \frac{F}{m_{\mathrm{A}} + m_{\mathrm{B}} + m_{\mathrm{C}}}$
### Step 1: Calculate Common Acceleration
a = frac805 + 3 + 2 = frac8010 = 8 mathrm~m/s^2$a = \frac{80}{5 + 3 + 2} = \frac{80}{10} = 8 \mathrm{~m/s}^2$
### Step 2: Calculate Tensions
Three masses (5 kg, 3 kg, 2 kg) connected by strings with tensions T1 and T2, being pulled by a common force F = 80N.mathrmT_1$\mathrm{T}_1$ pulls only block mathrmA$\mathrm{A}$ (5 mathrm~kg$5 \mathrm{~kg}$):
mathrmT_1 = m_mathrmA times a = 5 times 8 = 40 mathrm~N$\mathrm{T}_1 = m_{\mathrm{A}} \times a = 5 \times 8 = 40 \mathrm{~N}$Three masses (5 kg, 3 kg, 2 kg) connected by strings with tensions T1 and T2, being pulled by a common force F = 80N.mathrmT_2$\mathrm{T}_2$ pulls both blocks mathrmA$\mathrm{A}$ and mathrmB$\mathrm{B}$ (5 mathrm~kg + 3 mathrm~kg = 8 mathrm~kg$5 \mathrm{~kg} + 3 \mathrm{~kg} = 8 \mathrm{~kg}$):
From free body diagram of block B:
mathrmT_2 - mathrmT_1 = m_mathrmB times a$\mathrm{T}_2 - \mathrm{T}_1 = m_{\mathrm{B}} \times a$mathrmT_2 - 40 = 3 times 8 = 24 implies mathrmT_2 = 64 mathrm~N$\mathrm{T}_2 - 40 = 3 \times 8 = 24 \implies \mathrm{T}_2 = 64 \mathrm{~N}$
### 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
More Laws of Motion Questions — jee_main_2024_30_jan_morning
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