
A boy is pushing hard on a wall. Is he doing work?
(A) No
(B) Can’t say
(C) May have done some work
(D) Yes
Answer
232.8k+ views
Hint: We will see that when a boy pushes a wall, he is applying force on the wall. We see that the wall shows no displacement. Work is given as the product of the forces and the displacement in the same direction. Hence the work done is zero.
Complete step by step answer:
Work done:
Work is said to be done when force is applied on an object and it gets displaced from its initial position. SI unit of work is Joules.
1 Joule is defined as 1 newton force is applied on an object and it gets displaced by 1 m distance.
$\Rightarrow$ $1joule = 1\,Newton\,metres \Rightarrow 1\,J = 1\,Nm$
Mathematically;
$\Rightarrow$ $W = F\,s\,\cos \theta $ … (1)
W = work done
F = applied force
s = distance covered by an object
$\theta $= angle between applied force and object displacement
In this case when boy pushes on the wall, it shows no displacement
S = 0 … (2)
Using equation (2) in (1), we get
$W = F.0 \Rightarrow W = 0$
This represents that no work is done when force is applied on a wall.
Secondly, the wall does not get displaced from its initial position. So,
S = 0
Thus, the correct option is A. He is not doing any work.
Note:
The boy still gets tired even when the total mechanical work is zero. This means that the boy is using energy. To clarify this apparent paradox lets understand and Note: that energy needn’t be only converted to work; it can also get converted to other types of energies like heat, light, sound etc. IN our case the boy gets tired trying to oppose the rigidity of the wall. Theoretically the boy is compressing the wall. However, the wall’s tensile strength is too high hence the negligible amount of compression that the boy would have caused goes unnoticed. The boy’s energy is also lost in the form of heat due to which we would practically see him sweating
Complete step by step answer:
Work done:
Work is said to be done when force is applied on an object and it gets displaced from its initial position. SI unit of work is Joules.
1 Joule is defined as 1 newton force is applied on an object and it gets displaced by 1 m distance.
$\Rightarrow$ $1joule = 1\,Newton\,metres \Rightarrow 1\,J = 1\,Nm$
Mathematically;
$\Rightarrow$ $W = F\,s\,\cos \theta $ … (1)
W = work done
F = applied force
s = distance covered by an object
$\theta $= angle between applied force and object displacement
In this case when boy pushes on the wall, it shows no displacement
S = 0 … (2)
Using equation (2) in (1), we get
$W = F.0 \Rightarrow W = 0$
This represents that no work is done when force is applied on a wall.
Secondly, the wall does not get displaced from its initial position. So,
S = 0
Thus, the correct option is A. He is not doing any work.
Note:
The boy still gets tired even when the total mechanical work is zero. This means that the boy is using energy. To clarify this apparent paradox lets understand and Note: that energy needn’t be only converted to work; it can also get converted to other types of energies like heat, light, sound etc. IN our case the boy gets tired trying to oppose the rigidity of the wall. Theoretically the boy is compressing the wall. However, the wall’s tensile strength is too high hence the negligible amount of compression that the boy would have caused goes unnoticed. The boy’s energy is also lost in the form of heat due to which we would practically see him sweating
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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

