If a motor lifts the water
from a depth ‘h1’ and then
raises to a height ‘h2’ then
the power of the motor is
given by
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RGUIIIT BASAR
Slide 17
Power of heart = work/time
= (F.s)/t
= (PxA.s)/t
= (PxV )/t
= (hdgV)/t
(P = pressure,A = area of vessel, V = volume of vessel and
s = length of the vessel)
Thus power of heart = P(V/t) = (hdgV)/t
= Pressure x volume of blood pumped per second
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RGUIIIT BASAR
Slide 18
The heart of a man pumps 4 litres of blood per minute at a pressure of 130 m.m. of Hg. If the density of the blood is 13.6 gm/c.c. calculate the power of the heart.
Solution:-
Power of heart = (hdgV)/t
= 1.155watt.
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RGUIIIT BASAR
Slide 19
Power of lungs =
(mass of air blown per second ) x (velocity)2
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RGUIIIT BASAR
Slide 20
The work done by from time t1 to time t2 is given by
Where P = F.v
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RGUIIIT BASAR
W
Slide 21
If force F acted on a body of mass ‘m’ which is at rest , then the power produced in that body in time ‘t’ is given by
P = F.v
= F(F/m)t
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RGUIIIT BASAR
V = u + at
= 0 + at
= at
= (F/m)t
P
P = Fv
= mav
= ma(at)
= ma2t
Slide 22
A box of mass ‘m’ moved along a straight line by a machine delivering constant power(P).Then the distance moved by the body in terms of m, P & t is given by
Solution:- P=Fv = mav
P= m(dv/dt)v
vdv = (P/m)dt ,by integrating we get
dx/dt
On integrating we get,
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RGUIIIT BASAR
Slide 23
Electrical power
Instantaneous electrical power
The instantaneous electrical power P delivered to a component is given by
P(t) = V(t).I(t)
where P(t) is the instantaneous power, measured in watts (joules per second)
V(t) is the potential difference (or voltage drop) across the component, measured in volts
I(t) is the current through it, measured in amperes
If the component is a resistor, then:
P = V.I = I2 .R
where R is the resistance, measured in ohms.
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