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Question: Power supplied to a particle of mass \[2kg\] varies with time as \(P=\dfrac{3{{t}^{2}}}{2}watt\) , w...

Power supplied to a particle of mass 2kg2kg varies with time as P=3t22wattP=\dfrac{3{{t}^{2}}}{2}watt , where tt is in second. If the velocity of particle at t=0t=0 is v=0v=0, the velocity of particle at time t=2st=2s will be:
A.1m/sA.1m/s
B.4m/sB.4m/s
C.22m/sC.2\sqrt{2}m/s
D.2m/sD.2m/s

Explanation

Solution

We will use the work energy theorem to solve this problem. Work energy theorem relates work, power and energy. Work energy theorem states that work done by a net force on an object is equal to change in kinetic energy.
Formula Used:
We will use the following formula to solve this problem:-
t1t2Pdt=KEiKEf\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{Pdt}=K{{E}_{i}}-K{{E}_{f}} .

Complete answer:
From the question above we get the following parameters:-
It is being said that particles start from rest, it means that initial velocity, u=0m/su=0m/s .
Final velocity is considered as v=0m/sv=0m/s .
Time is from 0s0s to 2s2s . Hence, t1=0s{{t}_{1}}=0s and t2=2s{{t}_{2}}=2s .
Let the mass of the body be mm .
Now, using the following formula:-
t1t2Pdt=KEiKEf\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{Pdt}=K{{E}_{i}}-K{{E}_{f}} …………… (i)(i)
Putting values of the parameters in (i)(i) we get
023t22dt=12mv212mu2\int\limits_{0}^{2}{\dfrac{3{{t}^{2}}}{2}}dt=\dfrac{1}{2}m{{v}^{2}}-\dfrac{1}{2}m{{u}^{2}}
023t22dt=12mv20\int\limits_{0}^{2}{\dfrac{3{{t}^{2}}}{2}dt=\dfrac{1}{2}m{{v}^{2}}-0} ( As u=0m/su=0m/s )
32[t33]02=12×2×v2\Rightarrow \dfrac{3}{2}\left[ \dfrac{{{t}^{3}}}{3} \right]_{0}^{2}=\dfrac{1}{2}\times 2\times {{v}^{2}}
32×83=v2\Rightarrow \dfrac{3}{2}\times \dfrac{8}{3}={{v}^{2}}
v2=4\Rightarrow {{v}^{2}}=4
v=2m/s\Rightarrow v=2m/s
Hence, the velocity of particle at time t=2st=2s is given as v=2m/sv=2m/s.

Therefore, option (D)(D) is correct.

Additional Information:
Power is defined as the rate of doing work or energy consumed. SI unit of power is watt (W)(W) . According to the work energy theorem the work done on a body is equal to change in kinetic energy of a body. It is represented as W=KEfKEiW=K{{E}_{f}}-K{{E}_{i}} , where WW is work done, KEfK{{E}_{f}} is final kinetic energy and KEiK{{E}_{i}} is initial kinetic energy. Work done is due to all the forces on the body.

Note:
We know that work and energy are closely related to each other. It should be noted as work is also defined as the transfer of energy. It should also be noted that the work energy theorem is based on the law of conservation of energy.