Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How much power is needed to accelerate an object with a mass of $5kg$ and a velocity of $3m/s$ at a rate of $12m/{s^2}$ ?

Last updated date: 21st Jun 2024
Total views: 54.3k
Views today: 0.54k
Verified
54.3k+ 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.

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.