Column-1 (a) Block of mass 2 kg on a rough horizontal surface pulled by a horizontal force of 2

step 1 Hello everyone, let us do the following question. We have given two columns, column one and colu

Column-1 (a) Block of mass 2 kg on a rough horizontal...

Column-1 (a) Block of mass $2 \mathrm{~kg}$ on a rough horizontal surface pulled by a horizontal force of $20 \mathrm{~N}, \mu_{\mathrm{SL}}=0.5$ (b) Block of mass $2 \mathrm{~kg}$ pulled with constant speed up an incline of inclination $30^{\circ}$ and coeflicient of friction $\frac{1}{\sqrt{3}}$ (c) Block of mass $0.75 \mathrm{~kg}$ pulled by a constant [orce of $7.5 \mathrm{~N}$ upon inclinc of inclination $30^{\circ}$ and coefficient of friction $\frac{1}{\sqrt{3}}$ (d) Block of mass $2 \mathrm{~kg}$ pulled vertically by a force $20 \mathrm{~N}$ Column-lI (p) 'l'ension at the mid point of block is $10 \mathrm{~N}$ (q) \Lambdacceleration of block is $5 \mathrm{~m} / \mathrm{s}^{2}$ (r) Force of friction acting is $5 \mathrm{~N}$ (s) Resultant force on the block is zero

Column-1
(a) Block of mass $2 \mathrm{~kg}$ on a rough horizontal surface pulled by a horizontal force of $20 \mathrm{~N}, \mu_{\mathrm{SL}}=0.5$
(b) Block of mass $2 \mathrm{~kg}$ pulled with constant speed up an incline of inclination $30^{\circ}$ and coeflicient of friction $\frac{1}{\sqrt{3}}$
(c) Block of mass $0.75 \mathrm{~kg}$ pulled by a constant [orce of $7.5 \mathrm{~N}$ upon inclinc of inclination $30^{\circ}$ and coefficient of friction $\frac{1}{\sqrt{3}}$
(d) Block of mass $2 \mathrm{~kg}$ pulled vertically by a force $20 \mathrm{~N}$
Column-lI
(p) 'l'ension at the mid point of block is $10 \mathrm{~N}$
(q) \Lambdacceleration of block is $5 \mathrm{~m} / \mathrm{s}^{2}$
(r) Force of friction acting is $5 \mathrm{~N}$
(s) Resultant force on the block is zero

Key Concepts

Free Body Diagram Analysis

A free body diagram is a graphical representation of all the forces acting on a single object. It is an essential tool in mechanics that helps simplify complex problems by isolating the object and depicting forces such as applied forces, friction, weight, and normal reaction. This analysis is critical for setting up the equations needed to solve for unknown quantities in various physical scenarios.

Equilibrium Condition

The equilibrium condition refers to a state where the sum of all forces acting on an object is zero, resulting in no net acceleration. This concept is particularly important in problems where constant speed or static situations are encountered, allowing for the determination of unknown forces such as tension or friction by setting the net force to zero.

Friction

Friction is the resistive force that opposes the relative motion between two surfaces in contact. It is quantified by the coefficient of friction multiplied by the normal force, and can be classified as static or kinetic. Understanding friction is crucial for solving problems involving motion on horizontal surfaces, inclined planes, and in scenarios where objects begin to move or are in uniform motion.

Newton's Second Law

Newton’s Second Law, expressed as F = ma, is the fundamental principle used to determine the net force acting on an object and its resulting acceleration. This concept is essential in dynamics to relate the applied forces, frictional forces, gravitational forces, and any other forces present, providing a basis for calculating acceleration or proving equilibrium when the net force is zero.

Force Decomposition on Inclined Planes

Decomposing forces on inclined planes involves breaking down the gravitational force into components parallel and perpendicular to the plane. This technique is vital for analyzing motion on slopes, enabling the calculation of the net force acting along the incline and the corresponding frictional force, which in turn helps in determining acceleration or verifying equilibrium conditions.

Transcript

00:01 Hello everyone, let us do the following question.

00:03 We have given two columns, column one and column two.

00:06 We will solve this equation in the column one.

00:10 We have given a block of mass 2kg.

00:14 Here a horizontal is given.

00:16 Here a block is given of mass 2kg.

00:20 Here a force mg is acting vertical downward.

00:23 Reaction is acting upward.

00:25 And applied force of 20 newton is acting backward.

00:30 Frictional force of mu s m g is acting backward applied force is acting upward we have to tell which option? which option column 2 is correct? so here we will write that mass is given 2 kg applied force is given 20 newton static friction is given as 0 .5 newton so if we have to find the value of net force this will be be 20 minus mu m into g so value of this can be written as 20 mu can be written as 0 .5 in 0 .5 mass is 10 and g is 10 so net force is 10.

01:17 So this option is correct in column 2 where net force is given 10 newton.

01:23 Next we so tension at the midpoint is 10 newton.

01:26 This is correct.

01:27 Next we have to find acceleration.

01:30 Is given f upon m, force is 2, mass is 2, so acceleration is 5 meter per second scale.

01:39 Acceleration is given as 5 meter per second scale.

01:43 Next we have to find in column 1 a block is given of mass 2 kg, a block is given of mass 2 kg pulled with the constant speed.

01:56 Here it is given of mass 2 kg.

02:00 It is pulled up with the constant speed at angle of inclination 30 degree and coefficient of friction is given 1 by under the root 3.

02:10 Cofficient of friction is given 1 by under the root 3.

02:14 As in the question it is said that it moves up at constant speed, it moves up at constant speed.

02:23 So we will say net force on the block is 0.

02:28 So we will say net force on the block is zero...

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