Question

How much power is needed to accelerate an object with a mass of $4kg$ and a velocity of $8m{{s}^{-1}}$ at the rate of $6m{{s}^{-2}}$?

Answer 1

Hint: Work is being done on an object under the influence of a constant force as the acceleration is constant. Therefore, we can use equations of motion to calculate the time taken. Also calculate work done as change in energy. Power is the rate of doing work; substituting corresponding values in the above relation power can be calculated.
Formulas used:
$v=u+at$
$K=\dfrac{1}{2}m{{v}^{2}}$
$W=\Delta K={{K}_{2}}-{{K}_{1}}$
$P=\dfrac{W}{t}$

Complete answer:
Given that an object with mass $4kg$ is accelerated to a velocity $8m{{s}^{-1}}$ with an acceleration of $6m{{s}^{-2}}$.
Since the acceleration is constant, applying the following equation of motion we get,
$v=u+at$
Here, $v$ is the final velocity
$u$ is the initial velocity
$a$ is the acceleration
$t$ is the time taken
Assuming the object started from rest, we substitute given values in the above equation to get,
$\begin{align}
  & 8=0+6\times t \\
 & \Rightarrow t=1.33s \\
\end{align}$
Therefore, the object takes $1.33s$ to accelerate to the given velocity.
Change in kinetic energy of the object is equal to work done. Kinetic energy is given by-
$K=\dfrac{1}{2}m{{v}^{2}}$ - (1)
Here, $K$ is the kinetic energy
$m$ is the mass
$v$ is the velocity
Using eq (1), we get,
${{K}_{1}}=0$ - (2)
${{K}_{2}}=\dfrac{1}{2}\times 4\times {{(8)}^{2}}$
$\Rightarrow {{K}_{2}}=128J$ - (3)
Using eq (2) and eq (3), the change in kinetic energy will be-
$\begin{align}
  & W=\Delta K={{K}_{2}}-{{K}_{1}} \\
 & \Rightarrow W=128-0 \\
 & \therefore W=128J \\
\end{align}$
Therefore, the work done is $128J$.
Power is the rate at which work is done, therefore,
$P=\dfrac{W}{t}$
Given, $W=128J$, $t=1.33s$. Substituting given values in the above equation, we get,
$\begin{align}
  & P=\dfrac{128}{1.33} \\
 & \Rightarrow P=96.24J{{s}^{-1}} \\
\end{align}$
The power of the object is $96.24J{{s}^{-1}}$
Therefore, the power of the object accelerating at $6m{{s}^{-2}}$ is $96.24J{{s}^{-1}}$.

Note:
Power can also be defined as the product of force and velocity. Another method to solve this question is to calculate force as a product of mass and acceleration and use it to calculate power by substituting in the above relation. Power has a SI unit of Watt (W).