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**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.

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