
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
232.8k+ 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 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

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

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

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

