How much power is needed to accelerate an object with a mass of $5kg$ and a velocity of $3m/s$ at a rate of $12m/{s^2}$ ?
Answer
244.8k+ views
Hint: Studying the Work-Energy theorem will help us to solve the problem. According to the work-energy theorem, the NetWork on an object, plus the original kinetic energy is equal to the final kinetic energy of the object.
Complete step by step answer:
Work Energy theorem:
We know that mass$(m)$ of the object is $5kg, $velocity$(v)$ is$3m/s$ and acceleration$(a)$ is$12m/{s^2}$.
By using the equation of motion we can find the time taken by the object$(t)$which is accelerating,
$v = u + at.$ ($u$= initial velocity of the object which is obviously zero).
By substituting the value of $a$and $v$ we get,
$\Rightarrow 3 = 0 + 12t$
$ \Rightarrow t = \dfrac{3}{{12}}$
$ \Rightarrow t = \dfrac{1}{4}s$
Now by using the work-energy theorem, The net work performed on an object is proportional to the shift in kinetic energy of the object.
Assuming $W$ as work done by the object, $K.E{._f}$ as final kinetic energy, and $K.E{._i}$ as initial kinetic energy we calculate the work done as follows:
$\Rightarrow W = K.E{._f} - K.E{._i}$
$\Rightarrow (K.E. = \dfrac{1}{2}mv{}^2).$
$\Rightarrow W = \dfrac{1}{2}(5){(3)^2} - 0$ (Since the initial velocity is zero, the initial kinetic energy of the object is zero).
The work done is $W = \dfrac{{45}}{2}J$
The power needed is the rate of change of work done. The formula is derived as,
$\Rightarrow P = \dfrac{W}{t}$
$ \Rightarrow P = \dfrac{{45}}{2} \times 4$
Thus $P = 90Watts$
Hence by doing the above steps we find that the power needed by the object with the mass of $5kg$ to accelerate is equal to $90W.$
Note: Often people forget that only the net work, not the work performed by a particular power, is protected by the work-energy theorem. The law of work-energy states that the net work performed by the forces on an object is proportional to the difference in its kinetic energy.
Complete step by step answer:
Work Energy theorem:
We know that mass$(m)$ of the object is $5kg, $velocity$(v)$ is$3m/s$ and acceleration$(a)$ is$12m/{s^2}$.
By using the equation of motion we can find the time taken by the object$(t)$which is accelerating,
$v = u + at.$ ($u$= initial velocity of the object which is obviously zero).
By substituting the value of $a$and $v$ we get,
$\Rightarrow 3 = 0 + 12t$
$ \Rightarrow t = \dfrac{3}{{12}}$
$ \Rightarrow t = \dfrac{1}{4}s$
Now by using the work-energy theorem, The net work performed on an object is proportional to the shift in kinetic energy of the object.
Assuming $W$ as work done by the object, $K.E{._f}$ as final kinetic energy, and $K.E{._i}$ as initial kinetic energy we calculate the work done as follows:
$\Rightarrow W = K.E{._f} - K.E{._i}$
$\Rightarrow (K.E. = \dfrac{1}{2}mv{}^2).$
$\Rightarrow W = \dfrac{1}{2}(5){(3)^2} - 0$ (Since the initial velocity is zero, the initial kinetic energy of the object is zero).
The work done is $W = \dfrac{{45}}{2}J$
The power needed is the rate of change of work done. The formula is derived as,
$\Rightarrow P = \dfrac{W}{t}$
$ \Rightarrow P = \dfrac{{45}}{2} \times 4$
Thus $P = 90Watts$
Hence by doing the above steps we find that the power needed by the object with the mass of $5kg$ to accelerate is equal to $90W.$
Note: Often people forget that only the net work, not the work performed by a particular power, is protected by the work-energy theorem. The law of work-energy states that the net work performed by the forces on an object is proportional to the difference in its kinetic energy.
Recently Updated Pages
JEE Main 2026 Admit Card OUT LIVE Soon| Session 2 Direct Download Link

JEE Main 2026 Session 2 City Intimation Slip Expected Soon: Check How to Download

JEE Main 2026 Session 2 Application Form: Reopened Registration, Dates & Fees

JEE Main 2026 Session 2 Registration (Reopened): Last Date, Fees, Link & Process

WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

