JEE Main Previous Year Questions (2025): Friction | Physics for JEE Main & Advanced PDF Download

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JEE Main Previous Year Questions 
(2025): Friction 
Q1: A block of mass 1 kg , moving along ?? with speed ?? ?? = ???? ?? / ?? enters a rough region 
ranging from ?? = ?? . ?? ?? to ?? = ?? . ?? ?? . The retarding force acting on the block in this range is 
?? ?? = - ???? ?? , with k = ???? ?? /?? . Then the final speed of the block as it crosses rough region is. 
Options: 
A. 4 m / s 
B. 10 m / s 
C. 8 m / s 
D. 6 m / s 
Answer: C 
Solution: 
To find the final speed of a block moving through a rough region, consider the following steps: 
Given Parameters: 
Initial speed of the block, ?? ?? = 10 m / s. 
Mass of the block, ?? = 1 kg. 
Force of friction acting on the block, ?? ?? = - ?? · ?? . 
?? = 10 N / m. 
Forces and Acceleration: 
The acceleration ?? of the block is given by ?? =
?? ?? = - ???? . 
So, we have: ?? = - 10 ?? . 
Using the Relationship: 
We use the relationship between velocity, acceleration, and position: ?? ????
????
= ?? . 
Substituting ?? gives: ?? ????
????
= - 10 ?? . 
Integrating the Equation: 
Integrate both sides with respect to their variables: 
?
10
?? ? ???? ?? = - 10 ?
0 . 1
1 . 9
? ???? ?? 
Solving the Integrals: 
Page 2


JEE Main Previous Year Questions 
(2025): Friction 
Q1: A block of mass 1 kg , moving along ?? with speed ?? ?? = ???? ?? / ?? enters a rough region 
ranging from ?? = ?? . ?? ?? to ?? = ?? . ?? ?? . The retarding force acting on the block in this range is 
?? ?? = - ???? ?? , with k = ???? ?? /?? . Then the final speed of the block as it crosses rough region is. 
Options: 
A. 4 m / s 
B. 10 m / s 
C. 8 m / s 
D. 6 m / s 
Answer: C 
Solution: 
To find the final speed of a block moving through a rough region, consider the following steps: 
Given Parameters: 
Initial speed of the block, ?? ?? = 10 m / s. 
Mass of the block, ?? = 1 kg. 
Force of friction acting on the block, ?? ?? = - ?? · ?? . 
?? = 10 N / m. 
Forces and Acceleration: 
The acceleration ?? of the block is given by ?? =
?? ?? = - ???? . 
So, we have: ?? = - 10 ?? . 
Using the Relationship: 
We use the relationship between velocity, acceleration, and position: ?? ????
????
= ?? . 
Substituting ?? gives: ?? ????
????
= - 10 ?? . 
Integrating the Equation: 
Integrate both sides with respect to their variables: 
?
10
?? ? ???? ?? = - 10 ?
0 . 1
1 . 9
? ???? ?? 
Solving the Integrals: 
The left side becomes 
?? 2
2
-
100
2
. 
The right side evaluates to - 10 ·
( 1 . 9
2
- 0 . 1
2
)
2
. 
Therefore: 
?? 2
- 100
2
= - 10 (
1 . 9
2
- 0 . 1
2
2
) 
Final Calculation: 
Solve the above equation for ?? : 
?? 2
- 100
2
= - 10 ·
( 3 . 61 - 0 . 01 )
2
 
Simplifying and solving gives: 
?? 2
- 1 00 = - 10 · 1 . 8 
Solving further: 
?? 2
= 100 - 18 
?? 2
= 82 
?? = v 82 ˜ 8 m / s 
Therefore, the final speed of the block as it exits the rough region is approximately 8 m / s. 
 
Q2: A cubic block of mass ?? is sliding down on an inclined plane at ????
°
 with an acceleration 
of 
?? ?? , the value of coefficient of kinetic friction is 
JEE Main 2025 (Online) 7th April Morning Shift 
Options: 
A. 
v 3
2
 
B. 
v 2
3
 
C. 1 -
v 3
2
 
D. v 3 - 1 
Answer: D 
Solution: 
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JEE Main Previous Year Questions 
(2025): Friction 
Q1: A block of mass 1 kg , moving along ?? with speed ?? ?? = ???? ?? / ?? enters a rough region 
ranging from ?? = ?? . ?? ?? to ?? = ?? . ?? ?? . The retarding force acting on the block in this range is 
?? ?? = - ???? ?? , with k = ???? ?? /?? . Then the final speed of the block as it crosses rough region is. 
Options: 
A. 4 m / s 
B. 10 m / s 
C. 8 m / s 
D. 6 m / s 
Answer: C 
Solution: 
To find the final speed of a block moving through a rough region, consider the following steps: 
Given Parameters: 
Initial speed of the block, ?? ?? = 10 m / s. 
Mass of the block, ?? = 1 kg. 
Force of friction acting on the block, ?? ?? = - ?? · ?? . 
?? = 10 N / m. 
Forces and Acceleration: 
The acceleration ?? of the block is given by ?? =
?? ?? = - ???? . 
So, we have: ?? = - 10 ?? . 
Using the Relationship: 
We use the relationship between velocity, acceleration, and position: ?? ????
????
= ?? . 
Substituting ?? gives: ?? ????
????
= - 10 ?? . 
Integrating the Equation: 
Integrate both sides with respect to their variables: 
?
10
?? ? ???? ?? = - 10 ?
0 . 1
1 . 9
? ???? ?? 
Solving the Integrals: 
The left side becomes 
?? 2
2
-
100
2
. 
The right side evaluates to - 10 ·
( 1 . 9
2
- 0 . 1
2
)
2
. 
Therefore: 
?? 2
- 100
2
= - 10 (
1 . 9
2
- 0 . 1
2
2
) 
Final Calculation: 
Solve the above equation for ?? : 
?? 2
- 100
2
= - 10 ·
( 3 . 61 - 0 . 01 )
2
 
Simplifying and solving gives: 
?? 2
- 1 00 = - 10 · 1 . 8 
Solving further: 
?? 2
= 100 - 18 
?? 2
= 82 
?? = v 82 ˜ 8 m / s 
Therefore, the final speed of the block as it exits the rough region is approximately 8 m / s. 
 
Q2: A cubic block of mass ?? is sliding down on an inclined plane at ????
°
 with an acceleration 
of 
?? ?? , the value of coefficient of kinetic friction is 
JEE Main 2025 (Online) 7th April Morning Shift 
Options: 
A. 
v 3
2
 
B. 
v 2
3
 
C. 1 -
v 3
2
 
D. v 3 - 1 
Answer: D 
Solution: 
 
 
???? sin ? 60
°
- ?? ???? c os ? 60
°
= ???? 
?? sin ? 60 - ???? c os ? 60 =
?? 2
 
v 3
2
-
?? 2
=
1
2
 
?? = v 3 - 1