Question

The power supplied to the particle of mass 2 kg varies with the time as $\dfrac{3t^2}{2}$ watt where t is the time in seconds and the if the velocity of the particle at t = 0 is v = 0 then the velocity of the particle at t= 2 s will be 

A) $1\dfrac{m}{s}$

B) $4\dfrac{m}{s}$ 

C) $2\dfrac{m}{s} $

D) $2\sqrt{2}\dfrac{m}{s} $

Answer 1

Hint: Use the work-energy theorem and the definition of the power where we know 

The power is defined as the rate of energy transfer and hence the change in the energy can be obtained using the power given. The change in energy can be used to find the value of the final velocity.

Complete step by step solution:

The value of the power is changing with respect to the time and we know the definition of the power can be stated as the rate of the change of the energy 

$P = \dfrac{dE}{dt}$

So $\int P dt = \Delta E$  

So the integral of the power with respect to time gives us the change in the energy which we can consider as the change in the kinetic energy of the system. 

The value of the kinetic energy of the particle of the mass m is given as

$\dfrac{3t^2}{2}$ 

Where m is the mass of the particle and the value of the velocity is denoted using the term v 

The energy change can be obtained using the given values of the power as 

Change in KE  = $\int P dt$ 

The time interval given is : 0 to 2 sec

So integration gives the change in the kinetic energy of the system.

The value of the change in the kinetic energy of the system using the above method we get as

$\int \dfrac{3t^2}{2} dt = \dfrac{3}{2} (\dfrac{2^3- 0^2}{3} ) = 4 J$ 

So now we obtained the value of the change in the kinetic energy of the system as 4 J

The initial velocity of the particle is given as 0 and let's take the final value of the velocity as v 

So we get the value of the change in the kinetic energy in terms of the v as 

Change in KE = $\dfrac{1}{2} mv^2 - 0 = v^2$ 

Now this change in the kinetic energy is the same as we obtained earlier using the integration and we get the equation as below : 

$v^2 = 4$ and  $v = \sqrt 4 = 2 \dfrac{m}{s}$   

Thus we have used the concept of the power and found the value of the velocity of the particle after 2-second s as : $2 \dfrac{m}{s}$   

The correct answer is option (C).

Note: One of the main possible areas to make a mistake in this kind of problem is the integration part. We need to carefully integrate the power with respect to the time and find the value of the change in the energy by carefully applying the limits. If you take care of the integration,we can easily solve the problem using the power definition.