Answer
Verified
85.8k+ 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
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The thickness of the depletion layer is approximately class 11 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main