
A body of mass 5 Kg rets on a rough horizontal surface of coefficient of friction 0.2. The body is pulled through a distance of 10 m by a horizontal force of 25 N. Find the kinetic energy acquired by it.
(A) 200 J
(B) 150 J
(C) 100 J
(D) 50 J
Answer
161.7k+ views
Hint:First start with finding the resultant force on the body of mass 5 Kg given and then use the work energy theorem and find the work done on the given body which is equal to the resultant force multiplied by the distance moved by the body which is given as 10 m in the question.
Formula used:
Frictional force, ${F_k} = \mu R$
R is normal reaction on the body.
Work done, $W = {F_{net}} \times S$
Where, ${F_{net}}$ is resultant or net force
S is the distance .
Complete answer:
Normal reaction force on the body given is R =mg
$R = 5 \times 10 = 50N$
Horizontal force on the given body, ${F_H} = 25N$ (given)
Frictional force acting on the given body, ${F_k} = \mu R = 0.2 \times 50 = 10N$
Now, the net or resultant force on the body , ${F_{net}} = {F_H} - {F_k}$
${F_{net}} = 25 - 10 = 15N$
At last, work done will be:
$W = {F_{net}} \times S = 15 \times 10 = 150J$
$W = 150J$
Hence, the correct answer is Option(B).
Note: Be careful about the vertical and horizontal component of all the forces acting on the given body and find the value of the resultant or net force acting on the given body accordingly. Also check all the units of all the quantities given before putting the values in the required equation.
Formula used:
Frictional force, ${F_k} = \mu R$
R is normal reaction on the body.
Work done, $W = {F_{net}} \times S$
Where, ${F_{net}}$ is resultant or net force
S is the distance .
Complete answer:
Normal reaction force on the body given is R =mg
$R = 5 \times 10 = 50N$
Horizontal force on the given body, ${F_H} = 25N$ (given)
Frictional force acting on the given body, ${F_k} = \mu R = 0.2 \times 50 = 10N$
Now, the net or resultant force on the body , ${F_{net}} = {F_H} - {F_k}$
${F_{net}} = 25 - 10 = 15N$
At last, work done will be:
$W = {F_{net}} \times S = 15 \times 10 = 150J$
$W = 150J$
Hence, the correct answer is Option(B).
Note: Be careful about the vertical and horizontal component of all the forces acting on the given body and find the value of the resultant or net force acting on the given body accordingly. Also check all the units of all the quantities given before putting the values in the required equation.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
