Question

The power applied to a particle varies with time as P= 4t² - 3t+2 watt, where is in second. The change in its kinetic energy between time t= 1s to t= 3s plz answer me fast​

Answer 1

Answer:

$\boxed{\mathfrak{Change \ in \ kinetic \ energy \ (\Delta K) = 26.67 \ J}}$

Explanation:

Power w.r.t. time:

P = (4t² - 3t + 2) Watt

Rate of change of kinetic energy w.r.t. is known as power.

$\bf \dfrac{dK}{dt} = P \\ \bf \int\limits^{K_2}_{K_1} dK = \int\limits^{t_2}_{t_1} P.dt$

In the given question;

$\rm t_1 = 1 \ s$

$\rm t_2 = 3 \ s$

By substituting values in the equation we get:

$\\ \rm \implies \int\limits^{K_2}_{K_1} dK = \int\limits^{t_2 = 3 \: s}_{t_1 = 1 \: s} P.dt \\ \\ \rm \implies \int\limits^{K_2}_{K_1} dK = \int\limits^{3}_{1} (4 {t}^{2} - 3t + 2).dt \\ \\ \rm \implies K \big|_{K_1}^{K_2} = \big[ \dfrac{ 4{t}^{3} }{3} - \dfrac{ 3{t}^{2} }{2} + 2t \big]_{1}^{3} \\ \\ \rm \implies K_2 - K_1 = \dfrac{4( {3}^{3} - {1}^{3} )}{3} - \dfrac{3( {3}^{2} - {1}^{2}) }{2} + 2(3 - 1) \\ \\ \rm \implies \Delta K = \dfrac{4( 27 - 1 )}{3} - \dfrac{3( 9 - 1) }{2} + 2 \times 2 \\ \\ \rm \implies \Delta K = \dfrac{4 \times 26}{3} - \dfrac{3 \times 8}{2} + 4 \\ \\ \rm \implies \Delta K = \dfrac{104}{3} - (3 \times 4) + 4 \\ \\ \rm \implies \Delta K = \dfrac{104}{3} - 12 + 4 \\ \\ \rm \implies \Delta K = \dfrac{104}{3} - 8 \\ \\ \rm \implies \Delta K = \dfrac{104 - 24}{3} \\ \\ \rm \implies \Delta K = \dfrac{80}{3} \: J \\ \\ \rm \implies \Delta K = 26.67 \: J$

Answer 2

Power :

It is defined as change in energy per time .

$\orange{\bigstar}\ \; \rm P=\dfrac{dE}{dt}$

Solution :

P = 4t² - 3t + 2

We need to find change in K.E b/w t = 1 s to t = 3 s

Here it is given Kinetic energy , so formula changed to ,

$\green{\bigstar}\ \; \bf P=\dfrac{dK}{dt}$

$:\implies \rm dK=Pdt$

Integrating on both sides ,

$\displaystyle :\implies \rm K=\int Pdt$

$\displaystyle :\implies \rm K=\int_{1}^{3} (4t^2-3t+2)dt$

$:\implies \rm K=\left[\dfrac{4t^3}{3}-\dfrac{3t^2}{2}+2t \right]_1^3$

$:\implies \rm K=\left[\dfrac{4(3)^3}{3}-\dfrac{3(3)^2}{2}+2(3)\right]-\left[\dfrac{4(1)^3}{3}-\dfrac{3(1)^2}{2}+2(1)\right]$

$:\implies \rm K= \left[ 4(3)^2-\dfrac{3(9)}{2}+6\right]-\left[ \dfrac{4}{3}-\dfrac{3}{2}+2\right]$

$:\implies \rm K=\left[36-\dfrac{27}{2}+6\right]-\left[-\dfrac{1}{6}+2\right]$

$:\implies \rm K=40+\dfrac{1}{6}-\dfrac{27}{2}$

$:\implies \rm K=40-\dfrac{80}{6}$

$:\implies \rm K=\dfrac{160}{6}$

$:\implies \bf K=26.67\ J$

So , Change in Kinetic energy = 26.67 J