
A block of mass m slides along a surface where \[\mu \]is the coefficient of kinetic friction. A man pushes the block with a force F which makes an angle$\theta $with the horizontal as shown below. The block moves with constant velocity v. Power delivered by the man is:

A) $0$
B) $[F.v\cos \theta ]$
C) $[F\cos \theta - \mu (mg + F\sin \theta )]v$
D) $[F\cos \theta + \mu (mg + F\sin \theta )]v$$$
Answer
232.8k+ views
Hint: Recall that the frictional force is defined as the force that opposes the motion of the object when two surfaces are in contact with each other. The frictional force always acts in a direction opposite to the direction of motion of the object. It also varies with the type of surface of the object.
Complete step by step solution:
Since force is a vector quantity, it will have two components. One in horizontal direction which is $F\cos \theta $ and the other is vertical component which is $F\sin \theta $.
Also the normal force N will act in an upward direction and the weight ‘mg’ will act in a downward direction.
Also the coefficient of friction is given by the formula,
$\mu = \dfrac{F}{N}$
Where F is the frictional force
And N is the normal force
$F = \mu N$---(i)
The component of force along horizontal direction is $F\cos \theta $.
$\therefore $equation (i) can be written as
$F\cos \theta - \mu N = 0$---(ii)
The normal force is balanced by the vertical component of force and weight ‘mg’ as they are acting in an opposite direction to the force. Therefore, it can be written as
$\Rightarrow N = mg + F\sin \theta $---(iii)
Substituting the value of N from equation (iii) in equation (ii),
$\Rightarrow F\cos \theta - \mu [mg + F\sin \theta ] = 0$---(iv)
When an object is moving with a constant velocity with some applied force, the formula for power is written as
$Power = Force \times Velocity$
Substituting the value of force, from equation (iv)
$\Rightarrow P = [F\cos \theta - \mu (mg + F\sin \theta )]v$
The power delivered by the man is
$\Rightarrow P = [F\cos \theta - \mu (mg + F\sin \theta )]v$
Option (C) is the right answer.
Note: It is important to remember that the coefficient of friction is the ratio of frictional force and the normal force. When there is a lot of friction between the two surfaces, then the value of coefficient of friction is high. It measures how one object moves in relation to another object in contact.
Complete step by step solution:
Since force is a vector quantity, it will have two components. One in horizontal direction which is $F\cos \theta $ and the other is vertical component which is $F\sin \theta $.
Also the normal force N will act in an upward direction and the weight ‘mg’ will act in a downward direction.
Also the coefficient of friction is given by the formula,
$\mu = \dfrac{F}{N}$
Where F is the frictional force
And N is the normal force
$F = \mu N$---(i)
The component of force along horizontal direction is $F\cos \theta $.
$\therefore $equation (i) can be written as
$F\cos \theta - \mu N = 0$---(ii)
The normal force is balanced by the vertical component of force and weight ‘mg’ as they are acting in an opposite direction to the force. Therefore, it can be written as
$\Rightarrow N = mg + F\sin \theta $---(iii)
Substituting the value of N from equation (iii) in equation (ii),
$\Rightarrow F\cos \theta - \mu [mg + F\sin \theta ] = 0$---(iv)
When an object is moving with a constant velocity with some applied force, the formula for power is written as
$Power = Force \times Velocity$
Substituting the value of force, from equation (iv)
$\Rightarrow P = [F\cos \theta - \mu (mg + F\sin \theta )]v$
The power delivered by the man is
$\Rightarrow P = [F\cos \theta - \mu (mg + F\sin \theta )]v$
Option (C) is the right answer.
Note: It is important to remember that the coefficient of friction is the ratio of frictional force and the normal force. When there is a lot of friction between the two surfaces, then the value of coefficient of friction is high. It measures how one object moves in relation to another object in contact.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

