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
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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
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Answer:
Given:
Power applied to a particle is given as ;
[tex]P = 3 {t}^{2} - 2t + 1[/tex]
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:
[tex]work = \int \: P \: dt[/tex]
[tex] = > w= \int \: (3 {t}^{2} - 2t + 1) \: dt[/tex]
[tex] = > w = {t}^{3} - {t}^{2} + t[/tex]
Putting the limits :
[tex] = > w = \{{4}^{3} - {2}^{3} \} - \{{4}^{2} - {2}^{2} \} + \{4 - 2 \}[/tex]
[tex] = > w = 56 - 12 + 2[/tex]
[tex] = > w = 46 \: joule[/tex]
So , change in KE be :
[tex] = > \Delta KE \: = 46 \: joule[/tex]
So final answer :
[tex] \boxed{ \blue{ \sf{ \bold{ \huge{\Delta KE \: = 46 \: J}}}}}[/tex]
Answer:
[tex]\large\boxed{\sf{46\;Joules}}[/tex]
Explanation:
Given that, power applied to a particle varies with time as the relation,
[tex]p = 3 {t}^{2} - 2t + 1[/tex]
Where,
Now, we have to find the change in Kinetic Energy.
We know that, power is the rate of work done, i.e., [tex]\bold{\dfrac{dw}{dt}}[/tex].
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,
[tex] = > \dfrac{dw}{dt} = p \\ \\ = > dw = pdt[/tex]
Substituting the value of p, we get,
[tex] = > dw = (3 {t}^{2} - 2t + 1)dt[/tex]
Now, integrating both the sides, we get,
[tex] = > \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 [/tex]
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,
[tex] = > w_{ 2 \: sec. } = {2}^{3} - {2}^{2} + 2 \\ \\ = > w_{ 2 \: sec.} = 8 - 4 + 2 \\ \\ = > w_{ 2 \: sec.} = 6 \: joule[/tex]
Now, work done at t = 4 seconds is equal to,
[tex] = > w_{ 4 \: sec.} = {4}^{3} - {4}^{2} + 4 \\ \\ = > w_{ 4 \: sec.} = 64 - 16 + 4 \\ \\ = > w_{ 2 \: sec.} = 52 \: joule[/tex]
Therefore, change in kinetic energy is equal to,
[tex] = > \triangle K.E = w_{ 4\: sec.} - w_{ 2 \: sec.} \\ \\ = > \triangle K.E =( 52 - 6 )\: joule \\ \\ = > \triangle K.E = 46 \: joule[/tex]
Hence, change in kinetic energy is 46 Joules.