Question

37.A machine delivers power given by P=p0t2 0(t+t0)2P=p0t 02(t+t0)2 where p0p0 and t 0t 0 are constants. The machine starts at t=0t=0 and runs forever. What is maximum work that the machine can perform if p0p0 =3 SI unit, t 0t 0 =2 SI unit

Answer 1

Given :  A machine delivers power given by P =  p₀t₀²/(t + t₀)²  

p₀ , t₀ are constants

p₀ = 3

t₀ = 2

To Find : maximum work that the machine can perform

Solution:

dW =  Pdt

=> ∫dW = ∫Pdt

 $W=\int\limits^{\infty}_0 {P} \,dt$

 $W=\int\limits^{\infty}_0 {\dfrac{p_0t_0^2}{(t + t_0)^2}} \,dt$

 $W = \left [{-\dfrac{p_0t_0^2}{(t + t_0)}}\right] ^{\infty}_0$

 $W = 0 -\left(-\dfrac{p_0t_0^2}{(0 + t_0)}\right)$

$W = \dfrac{p_0t_0^2}{( t_0)}$

W = p₀t₀

p₀ = 3

t₀ = 2

W = 3 * 2

=> W = 6

maximum 6 SI unit work can be done

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