A pump motor is used to deliver water at a certain rate from a given pipe. To obtain “2” time water from the same pipe in the same time, the amount to which the power of the motor should be increased is :
Force on water coming out is F= ρAv² and Power P=F.v
So, Power = F.v= ρAv³,
if v becomes 2v, Power P1= 8P
If Potential energy is given by U = (x³ + y³ + z²). Calculate force at a point (3, 2, 1) :
A position dependent force, \(F = (7 – 2x + 3x^2)\text{ N})\) acts on a small body of mass \(2\text{ kg})\) and displaces it from \(x = 0\) to \(x = 5\text{ m})\). The work done in joule is:
(1994, 92)
\(W = int F dx\). Given \(F = 7 - 2x + 3x^2\). Limits \(x = 0\) to \(x = 5\text{ m})\). \(W = [7x - x^2 + x^3]_0^5 = 7(5) - 5^2 + 5^3 = 35 - 25 + 125 = 135\text{ J}\).