JEE Main Previous Year Questions (2025): Work, Energy & Power

 Page 1


JEE Main Previous Year Qs (2025): 
Work, Energy & Power 
 
Q1: A force ?? = ?? ?? ?? ??ˆ + ?? ?? ??ˆ acts on a particle in a plane ?? + ?? = ???? . The work done by 
this force during a displacement from (?? ,?? ) to (?? ?? ,?? ?? ) is ____ Joule (round off to 
the nearest integer) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 152 
Solution: 
To calculate the work done by the force ?? = ?? 2
?? ??ˆ + ?? 2
??ˆ during the displacement from (0,0) to 
(4 m,2 m) , we proceed as follows: 
Given the constraint that the particle moves in the plane x + y = 10, we can express y in terms 
of x : 
?? = 10- ?? 
The work done, ?? , is the integral of the force along the path of the displacement: 
?? = ?
0
4
??? 2
(10- ?? )???? + ?
0
2
??? 2
???? 
Calculate the integral of the i ˆ component: 
 ? ?
4
0
??? 2
(10- ?? )???? = ? ?
4
0
?(10?? 2
- ?? 3
)????
 = [
10?? 3
3
-
?? 4
4
]
0
4
 = (
10(4)
3
3
-
(4)
4
4
) - (
10(0)
3
3
-
(0)
4
4
)
 =
640
3
-
256
4
 
Calculate the integral of the j ˆ component: 
 ? ?
2
0
??? 2
???? = [
?? 3
3
]
0
2
 =
(2)
3
3
-
(0)
3
3
 =
8
3
 
Combine the results: 
?? = (
640
3
- 64)+
8
3
 
Page 2


JEE Main Previous Year Qs (2025): 
Work, Energy & Power 
 
Q1: A force ?? = ?? ?? ?? ??ˆ + ?? ?? ??ˆ acts on a particle in a plane ?? + ?? = ???? . The work done by 
this force during a displacement from (?? ,?? ) to (?? ?? ,?? ?? ) is ____ Joule (round off to 
the nearest integer) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 152 
Solution: 
To calculate the work done by the force ?? = ?? 2
?? ??ˆ + ?? 2
??ˆ during the displacement from (0,0) to 
(4 m,2 m) , we proceed as follows: 
Given the constraint that the particle moves in the plane x + y = 10, we can express y in terms 
of x : 
?? = 10- ?? 
The work done, ?? , is the integral of the force along the path of the displacement: 
?? = ?
0
4
??? 2
(10- ?? )???? + ?
0
2
??? 2
???? 
Calculate the integral of the i ˆ component: 
 ? ?
4
0
??? 2
(10- ?? )???? = ? ?
4
0
?(10?? 2
- ?? 3
)????
 = [
10?? 3
3
-
?? 4
4
]
0
4
 = (
10(4)
3
3
-
(4)
4
4
) - (
10(0)
3
3
-
(0)
4
4
)
 =
640
3
-
256
4
 
Calculate the integral of the j ˆ component: 
 ? ?
2
0
??? 2
???? = [
?? 3
3
]
0
2
 =
(2)
3
3
-
(0)
3
3
 =
8
3
 
Combine the results: 
?? = (
640
3
- 64)+
8
3
 
Simplify: 
?? =
640
3
- 64+
8
3
 =
640+ 8
3
- 64
 =
648
3
- 64
 = 216- 64
 = 152 J
 
Thus, the work done by the force is 152 Joules. 
Q2: In a hydraulic lift, the surface area of the input piston is ?? ????
?? and that of the 
output piston is ???????? ????
?? . If 100 N force is applied to the input piston to raise the 
output piston by 20 cm , then the work done is ____ kJ. 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 5 
Solution: 
 
F
1
 A
1
=
F
2
?? 2
,
100
6
=
F
1500
, F =
50
3
× 1500
 F = 50× 500 = 25× 10
3
 N
?? = F
? 
· S
?? 
= 25× 10
3
×
20
100
 = 5 × 10
3
= 5 kJ
 
Q3: A force ?? 
= ?? ??ˆ + ?? ??ˆ + ?? ˆ
 is applied on a particle and it undergoes a displacement 
??ˆ - ?? ??ˆ - ?? ˆ
 What will be the value of ?? , if work done on the particle is zero. 
Page 3


JEE Main Previous Year Qs (2025): 
Work, Energy & Power 
 
Q1: A force ?? = ?? ?? ?? ??ˆ + ?? ?? ??ˆ acts on a particle in a plane ?? + ?? = ???? . The work done by 
this force during a displacement from (?? ,?? ) to (?? ?? ,?? ?? ) is ____ Joule (round off to 
the nearest integer) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 152 
Solution: 
To calculate the work done by the force ?? = ?? 2
?? ??ˆ + ?? 2
??ˆ during the displacement from (0,0) to 
(4 m,2 m) , we proceed as follows: 
Given the constraint that the particle moves in the plane x + y = 10, we can express y in terms 
of x : 
?? = 10- ?? 
The work done, ?? , is the integral of the force along the path of the displacement: 
?? = ?
0
4
??? 2
(10- ?? )???? + ?
0
2
??? 2
???? 
Calculate the integral of the i ˆ component: 
 ? ?
4
0
??? 2
(10- ?? )???? = ? ?
4
0
?(10?? 2
- ?? 3
)????
 = [
10?? 3
3
-
?? 4
4
]
0
4
 = (
10(4)
3
3
-
(4)
4
4
) - (
10(0)
3
3
-
(0)
4
4
)
 =
640
3
-
256
4
 
Calculate the integral of the j ˆ component: 
 ? ?
2
0
??? 2
???? = [
?? 3
3
]
0
2
 =
(2)
3
3
-
(0)
3
3
 =
8
3
 
Combine the results: 
?? = (
640
3
- 64)+
8
3
 
Simplify: 
?? =
640
3
- 64+
8
3
 =
640+ 8
3
- 64
 =
648
3
- 64
 = 216- 64
 = 152 J
 
Thus, the work done by the force is 152 Joules. 
Q2: In a hydraulic lift, the surface area of the input piston is ?? ????
?? and that of the 
output piston is ???????? ????
?? . If 100 N force is applied to the input piston to raise the 
output piston by 20 cm , then the work done is ____ kJ. 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 5 
Solution: 
 
F
1
 A
1
=
F
2
?? 2
,
100
6
=
F
1500
, F =
50
3
× 1500
 F = 50× 500 = 25× 10
3
 N
?? = F
? 
· S
?? 
= 25× 10
3
×
20
100
 = 5 × 10
3
= 5 kJ
 
Q3: A force ?? 
= ?? ??ˆ + ?? ??ˆ + ?? ˆ
 is applied on a particle and it undergoes a displacement 
??ˆ - ?? ??ˆ - ?? ˆ
 What will be the value of ?? , if work done on the particle is zero. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
1
3
 
B. 
1
2
 
C. 0 
D. 2 
Ans: B 
Solution: 
To determine the value of ?? such that the work done on the particle is zero, follow these steps: 
The work done by a force on a displacement is given by the dot product: 
Work = ?? 
· ?? 
. 
Given the force 
?? 
= 2??ˆ + ?? ??ˆ + ?? ˆ
, 
and the displacement 
?? 
= ??ˆ - 2??ˆ - ?? ˆ
, 
compute the dot product: 
?? 
· ?? 
= (2)(1)+ (?? )(-2)+ (1)(-1) . 
Simplify the expression: 
?? 
· ?? 
= 2- 2?? - 1 = 1 - 2?? . 
Since the work done is zero: 
1 - 2?? = 0. 
Solve for ?? : 
2?? = 1, 
?? =
1
2
. 
Thus, the value of ?? is 
1
2
, which corresponds to Option B. 
Q4: A ball having kinetic energy KE, is projected at an angle of ????
°
 from the horizontal. 
What will be the kinetic energy of ball at the highest point of its flight? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 
(KE)
2
 
B. 
(KE)
8
 
C. 
(KE)
4
 
D. 
(KE)
16
 
Page 4


JEE Main Previous Year Qs (2025): 
Work, Energy & Power 
 
Q1: A force ?? = ?? ?? ?? ??ˆ + ?? ?? ??ˆ acts on a particle in a plane ?? + ?? = ???? . The work done by 
this force during a displacement from (?? ,?? ) to (?? ?? ,?? ?? ) is ____ Joule (round off to 
the nearest integer) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 152 
Solution: 
To calculate the work done by the force ?? = ?? 2
?? ??ˆ + ?? 2
??ˆ during the displacement from (0,0) to 
(4 m,2 m) , we proceed as follows: 
Given the constraint that the particle moves in the plane x + y = 10, we can express y in terms 
of x : 
?? = 10- ?? 
The work done, ?? , is the integral of the force along the path of the displacement: 
?? = ?
0
4
??? 2
(10- ?? )???? + ?
0
2
??? 2
???? 
Calculate the integral of the i ˆ component: 
 ? ?
4
0
??? 2
(10- ?? )???? = ? ?
4
0
?(10?? 2
- ?? 3
)????
 = [
10?? 3
3
-
?? 4
4
]
0
4
 = (
10(4)
3
3
-
(4)
4
4
) - (
10(0)
3
3
-
(0)
4
4
)
 =
640
3
-
256
4
 
Calculate the integral of the j ˆ component: 
 ? ?
2
0
??? 2
???? = [
?? 3
3
]
0
2
 =
(2)
3
3
-
(0)
3
3
 =
8
3
 
Combine the results: 
?? = (
640
3
- 64)+
8
3
 
Simplify: 
?? =
640
3
- 64+
8
3
 =
640+ 8
3
- 64
 =
648
3
- 64
 = 216- 64
 = 152 J
 
Thus, the work done by the force is 152 Joules. 
Q2: In a hydraulic lift, the surface area of the input piston is ?? ????
?? and that of the 
output piston is ???????? ????
?? . If 100 N force is applied to the input piston to raise the 
output piston by 20 cm , then the work done is ____ kJ. 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 5 
Solution: 
 
F
1
 A
1
=
F
2
?? 2
,
100
6
=
F
1500
, F =
50
3
× 1500
 F = 50× 500 = 25× 10
3
 N
?? = F
? 
· S
?? 
= 25× 10
3
×
20
100
 = 5 × 10
3
= 5 kJ
 
Q3: A force ?? 
= ?? ??ˆ + ?? ??ˆ + ?? ˆ
 is applied on a particle and it undergoes a displacement 
??ˆ - ?? ??ˆ - ?? ˆ
 What will be the value of ?? , if work done on the particle is zero. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
1
3
 
B. 
1
2
 
C. 0 
D. 2 
Ans: B 
Solution: 
To determine the value of ?? such that the work done on the particle is zero, follow these steps: 
The work done by a force on a displacement is given by the dot product: 
Work = ?? 
· ?? 
. 
Given the force 
?? 
= 2??ˆ + ?? ??ˆ + ?? ˆ
, 
and the displacement 
?? 
= ??ˆ - 2??ˆ - ?? ˆ
, 
compute the dot product: 
?? 
· ?? 
= (2)(1)+ (?? )(-2)+ (1)(-1) . 
Simplify the expression: 
?? 
· ?? 
= 2- 2?? - 1 = 1 - 2?? . 
Since the work done is zero: 
1 - 2?? = 0. 
Solve for ?? : 
2?? = 1, 
?? =
1
2
. 
Thus, the value of ?? is 
1
2
, which corresponds to Option B. 
Q4: A ball having kinetic energy KE, is projected at an angle of ????
°
 from the horizontal. 
What will be the kinetic energy of ball at the highest point of its flight? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 
(KE)
2
 
B. 
(KE)
8
 
C. 
(KE)
4
 
D. 
(KE)
16
 
Ans: C 
Solution: 
Let's break down the problem step by step: 
The ball is projected with kinetic energy 
KE=
1
2
?? ?? 2
, 
where ?? is the initial speed. 
The ball is launched at an angle of 60
°
. So, its initial velocity components are: 
Horizontal: ?? ?? = ?? cos (60
°
)=
?? 2
 
Vertical: ?? ?? = ?? sin (60
°
) 
At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., ?? ?? =
0 ), but the horizontal component remains unchanged. 
Therefore, the kinetic energy at the highest point is only due to the horizontal velocity: 
KE
top 
=
1
2
?? ?? ?? 2
 
Substitute ?? ?? =
?? 2
 : 
KE
top 
=
1
2
?? (
?? 2
)
2
=
1
2
?? ?? 2
4
=
1
8
?? ?? 2
 
Notice that the initial kinetic energy is 
KE=
1
2
?? ?? 2
. 
So we can write: 
KE
top 
=
1
8
?? ?? 2
=
1
4
(
1
2
?? ?? 2
) =
1
4
KE 
Thus, the kinetic energy at the highest point is 
KE
4
. 
Ans: Option C. 
Q5: A force ?? = ?? + ?? ?? ?? acts on an object in the x -direction. The work done by the 
force is 5 J when the object is displaced by 1 m . If the constant ?? = ?? ?? then ?? will be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 15 N/m
2
 
B. 10 N/m
2
 
C. 12 N/m
2
 
D. 8 N/m
2
 
Ans: C 
Solution: 
Page 5


JEE Main Previous Year Qs (2025): 
Work, Energy & Power 
 
Q1: A force ?? = ?? ?? ?? ??ˆ + ?? ?? ??ˆ acts on a particle in a plane ?? + ?? = ???? . The work done by 
this force during a displacement from (?? ,?? ) to (?? ?? ,?? ?? ) is ____ Joule (round off to 
the nearest integer) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Ans: 152 
Solution: 
To calculate the work done by the force ?? = ?? 2
?? ??ˆ + ?? 2
??ˆ during the displacement from (0,0) to 
(4 m,2 m) , we proceed as follows: 
Given the constraint that the particle moves in the plane x + y = 10, we can express y in terms 
of x : 
?? = 10- ?? 
The work done, ?? , is the integral of the force along the path of the displacement: 
?? = ?
0
4
??? 2
(10- ?? )???? + ?
0
2
??? 2
???? 
Calculate the integral of the i ˆ component: 
 ? ?
4
0
??? 2
(10- ?? )???? = ? ?
4
0
?(10?? 2
- ?? 3
)????
 = [
10?? 3
3
-
?? 4
4
]
0
4
 = (
10(4)
3
3
-
(4)
4
4
) - (
10(0)
3
3
-
(0)
4
4
)
 =
640
3
-
256
4
 
Calculate the integral of the j ˆ component: 
 ? ?
2
0
??? 2
???? = [
?? 3
3
]
0
2
 =
(2)
3
3
-
(0)
3
3
 =
8
3
 
Combine the results: 
?? = (
640
3
- 64)+
8
3
 
Simplify: 
?? =
640
3
- 64+
8
3
 =
640+ 8
3
- 64
 =
648
3
- 64
 = 216- 64
 = 152 J
 
Thus, the work done by the force is 152 Joules. 
Q2: In a hydraulic lift, the surface area of the input piston is ?? ????
?? and that of the 
output piston is ???????? ????
?? . If 100 N force is applied to the input piston to raise the 
output piston by 20 cm , then the work done is ____ kJ. 
JEE Main 2025 (Online) 29th January Morning Shift 
Ans: 5 
Solution: 
 
F
1
 A
1
=
F
2
?? 2
,
100
6
=
F
1500
, F =
50
3
× 1500
 F = 50× 500 = 25× 10
3
 N
?? = F
? 
· S
?? 
= 25× 10
3
×
20
100
 = 5 × 10
3
= 5 kJ
 
Q3: A force ?? 
= ?? ??ˆ + ?? ??ˆ + ?? ˆ
 is applied on a particle and it undergoes a displacement 
??ˆ - ?? ??ˆ - ?? ˆ
 What will be the value of ?? , if work done on the particle is zero. 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
1
3
 
B. 
1
2
 
C. 0 
D. 2 
Ans: B 
Solution: 
To determine the value of ?? such that the work done on the particle is zero, follow these steps: 
The work done by a force on a displacement is given by the dot product: 
Work = ?? 
· ?? 
. 
Given the force 
?? 
= 2??ˆ + ?? ??ˆ + ?? ˆ
, 
and the displacement 
?? 
= ??ˆ - 2??ˆ - ?? ˆ
, 
compute the dot product: 
?? 
· ?? 
= (2)(1)+ (?? )(-2)+ (1)(-1) . 
Simplify the expression: 
?? 
· ?? 
= 2- 2?? - 1 = 1 - 2?? . 
Since the work done is zero: 
1 - 2?? = 0. 
Solve for ?? : 
2?? = 1, 
?? =
1
2
. 
Thus, the value of ?? is 
1
2
, which corresponds to Option B. 
Q4: A ball having kinetic energy KE, is projected at an angle of ????
°
 from the horizontal. 
What will be the kinetic energy of ball at the highest point of its flight? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 
(KE)
2
 
B. 
(KE)
8
 
C. 
(KE)
4
 
D. 
(KE)
16
 
Ans: C 
Solution: 
Let's break down the problem step by step: 
The ball is projected with kinetic energy 
KE=
1
2
?? ?? 2
, 
where ?? is the initial speed. 
The ball is launched at an angle of 60
°
. So, its initial velocity components are: 
Horizontal: ?? ?? = ?? cos (60
°
)=
?? 2
 
Vertical: ?? ?? = ?? sin (60
°
) 
At the highest point of its flight, the vertical component of the velocity becomes zero (i.e., ?? ?? =
0 ), but the horizontal component remains unchanged. 
Therefore, the kinetic energy at the highest point is only due to the horizontal velocity: 
KE
top 
=
1
2
?? ?? ?? 2
 
Substitute ?? ?? =
?? 2
 : 
KE
top 
=
1
2
?? (
?? 2
)
2
=
1
2
?? ?? 2
4
=
1
8
?? ?? 2
 
Notice that the initial kinetic energy is 
KE=
1
2
?? ?? 2
. 
So we can write: 
KE
top 
=
1
8
?? ?? 2
=
1
4
(
1
2
?? ?? 2
) =
1
4
KE 
Thus, the kinetic energy at the highest point is 
KE
4
. 
Ans: Option C. 
Q5: A force ?? = ?? + ?? ?? ?? acts on an object in the x -direction. The work done by the 
force is 5 J when the object is displaced by 1 m . If the constant ?? = ?? ?? then ?? will be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 15 N/m
2
 
B. 10 N/m
2
 
C. 12 N/m
2
 
D. 8 N/m
2
 
Ans: C 
Solution: 
The work done by a force is given by the integral of the force over the displacement. The force is 
given by: 
?? (?? )= ?? + ?? ?? 2
 
To find the work done when the object is displaced from ?? = 0 to ?? = 1, we compute: 
?? = ?
0
1
?(?? + ?? ?? 2
)???? 
Substituting ?? = 1 N : 
?? = ?
0
1
?(1+ ?? ?? 2
)???? = ?
0
1
?1???? + ?
0
1
??? ?? 2
???? 
Calculating each integral separately: 
?
0
1
?1???? = [?? ]
0
1
= 1 
?
0
1
??? ?? 2
???? = ?? [
?? 3
3
]
0
1
= ?? [
1
3
3
-
0
3
3
] =
?? 3
 
Thus, the total work done is: 
?? = 1 +
?? 3
 
We are given that the work done is 5 J , so: 
1 +
?? 3
= 5 
Subtract 1 from both sides: 
?? 3
= 4 
Multiply both sides by 3 to solve for ?? : 
?? = 12 N/m
2
 
Therefore, the value of ?? is 12 N/m
2
. Thus, Option C 12 N/m
2
 is the correct answer. 
Q6: Given below are two statements: one is labelled as Assertion A and the other is 
labelled as Reason ?? 
Assertion A: In a central force field, the work done is independent of the path chosen. 
Reason R: Every force encountered in mechanics does not have an associated 
potential energy. 
In the light of the above statements, choose the most appropriate answer from the 
options given below 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. ?? is false but ?? is true 
B. Both ?? and ?? are true but ?? is NOT the correct explanation of ?? 
C. ?? is true but ?? is false 
D. Both ?? and ?? are true and ?? is the correct explanation of ?? 
Ans: B 
Solution: