
A force of 5N acts on a 15kg body initially at rest. The work done by the force during the first second of motion of the body is
A. 5J
B. $\dfrac{5}{5}$ J
C. 6J
D. 75J
Answer
162.6k+ views
Hint: A body’s state of motion or rest can be changed by the application of an external force. In other words, force is a physical push or pull exerted on a body having a definite mass which causes the body to change its velocity and opposes its inertia. Inertia on the other hand can be defined as the inherent tendency of an object to continue to stay in its state of motion or rest. When a force moves an object over a distance, "work" has been said to be done.
Formula used:
Expression of force is,
$F=ma$
Where, $F$ is the force, $a$ is the acceleration and $m$ is the mass.
Equation of motion is,
$S = ut + \dfrac{1}{2}\,a{t^2}$
Where, $S$ is displacement, $u$ is the initial velocity and $t$ is the time.
Complete step by step solution:
We know, Force, F = mass (m) \[ \times \] acceleration (a).
So, in the question given, F= 5N and mass = 15Kg,
Putting the above values in the equation of force, we get:
5 = 15 \[ \times \] a
\[ \Rightarrow a = \dfrac{1}{3}\,m{s^{ - 2}}\]
Now, we are aware of the second equation of motion i.e. $S = ut + \dfrac{1}{2}\,a{t^2}$
Putting the values of a and t in this equation with time(t) being 1 sec as given in the question, we get:
\[S = 0 + \dfrac{1}{2}\,\dfrac{1}{3}\,\times {1^2}\]
\[ \Rightarrow S = \dfrac{1}{6}\]
The formula for work done is given by, Work = Force \[ \times \] displacement produced
Substituting the value of “S” in place of the displacement in the formula, we get:
W = 5 \[ \times \dfrac{1}{6}\]
\[ \Rightarrow \]W = \dfrac{5}{6}\] J
Hence, overall work done in moving the 15Kg body with a force of 5N in its first second is $\dfrac{5}{6}$ Joules.
So, option B will be the appropriate response to this question .
Note: Always remember to use proper units while solving these types of problems. Convert the units if needed to find the correct answer. Proper conversion and use of S.I and C.G.S systems of units are often required to find and maintain the proper unit in the answer. Always remember that the lowercase “s” stands for seconds and the uppercase “S” stands for displacement.
Formula used:
Expression of force is,
$F=ma$
Where, $F$ is the force, $a$ is the acceleration and $m$ is the mass.
Equation of motion is,
$S = ut + \dfrac{1}{2}\,a{t^2}$
Where, $S$ is displacement, $u$ is the initial velocity and $t$ is the time.
Complete step by step solution:
We know, Force, F = mass (m) \[ \times \] acceleration (a).
So, in the question given, F= 5N and mass = 15Kg,
Putting the above values in the equation of force, we get:
5 = 15 \[ \times \] a
\[ \Rightarrow a = \dfrac{1}{3}\,m{s^{ - 2}}\]
Now, we are aware of the second equation of motion i.e. $S = ut + \dfrac{1}{2}\,a{t^2}$
Putting the values of a and t in this equation with time(t) being 1 sec as given in the question, we get:
\[S = 0 + \dfrac{1}{2}\,\dfrac{1}{3}\,\times {1^2}\]
\[ \Rightarrow S = \dfrac{1}{6}\]
The formula for work done is given by, Work = Force \[ \times \] displacement produced
Substituting the value of “S” in place of the displacement in the formula, we get:
W = 5 \[ \times \dfrac{1}{6}\]
\[ \Rightarrow \]W = \dfrac{5}{6}\] J
Hence, overall work done in moving the 15Kg body with a force of 5N in its first second is $\dfrac{5}{6}$ Joules.
So, option B will be the appropriate response to this question .
Note: Always remember to use proper units while solving these types of problems. Convert the units if needed to find the correct answer. Proper conversion and use of S.I and C.G.S systems of units are often required to find and maintain the proper unit in the answer. Always remember that the lowercase “s” stands for seconds and the uppercase “S” stands for displacement.
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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Charging and Discharging of Capacitor

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
