Question

wer
Power applied to a particle varies with time
as P=(3t2-2t+1) watt, where tis in second. The
change in its kinetic energy between time t#2
sec. and t=4 sec. is
1) 32 J 2) 46J 3) 61
J 4 ) 102 J​

Answer 1

Answer:

Given:

Power applied to a particle is given as ;

$P = 3 {t}^{2} - 2t + 1$

To find:

Work done in the specified time interval.

Concept:

Power is the rate of doing work. In other words , it's the ratio of work done to the time taken.

On the other hand , as per Work Energy Theorem, we can say that the work done by all the forces will be equal to the Change in Kinetic Energy.

Calculation:

$work = \int \: P \: dt$

$= > w= \int \: (3 {t}^{2} - 2t + 1) \: dt$

$= > w = {t}^{3} - {t}^{2} + t$

Putting the limits :

$= > w = \{{4}^{3} - {2}^{3} \} - \{{4}^{2} - {2}^{2} \} + \{4 - 2 \}$

$= > w = 56 - 12 + 2$

$= > w = 46 \: joule$

So , change in KE be :

$= > \Delta KE \: = 46 \: joule$

So final answer :

$\boxed{ \blue{ \sf{ \bold{ \huge{\Delta KE \: = 46 \: J}}}}}$

Answer 2

Answer:

$\large\boxed{\sf{46\;Joules}}$

Explanation:

Given that, power applied to a particle varies with time as the relation,

$p = 3 {t}^{2} - 2t + 1$

Where,

• p is power applied in Watt
• t is time in seconds

Now, we have to find the change in Kinetic Energy.

We know that, power is the rate of work done, i.e., $\bold{\dfrac{dw}{dt}}$.

Also, by work energy theorem, we know that, total work done is equals to change in kinetic energy.

Therefore, let's find the work done.

Therefore, we will get,

$= > \dfrac{dw}{dt} = p \\ \\ = > dw = pdt$

Substituting the value of p, we get,

$= > dw = (3 {t}^{2} - 2t + 1)dt$

Now, integrating both the sides, we get,

$= > \displaystyle\int \: dw = \displaystyle\int (3 {t}^{2} - 2t + 1)dt \\ \\ = > w = 3 \frac{ {t}^{3} }{3} - 2 \frac{ {t}^{2} }{2} + t + c\\ \\ = > w = {t}^{ 3} - {t}^{2} + t$

Now, we have to find the work done between time t = 2 and t = 4 second.

Now, work done at t = 2 seconds is equal to,

$= > w_{ 2 \: sec. } = {2}^{3} - {2}^{2} + 2 \\ \\ = > w_{ 2 \: sec.} = 8 - 4 + 2 \\ \\ = > w_{ 2 \: sec.} = 6 \: joule$

Now, work done at t = 4 seconds is equal to,

$= > w_{ 4 \: sec.} = {4}^{3} - {4}^{2} + 4 \\ \\ = > w_{ 4 \: sec.} = 64 - 16 + 4 \\ \\ = > w_{ 2 \: sec.} = 52 \: joule$

Therefore, change in kinetic energy is equal to,

$= > \triangle K.E = w_{ 4\: sec.} - w_{ 2 \: sec.} \\ \\ = > \triangle K.E =( 52 - 6 )\: joule \\ \\ = > \triangle K.E = 46 \: joule$

Hence, change in kinetic energy is 46 Joules.