An engine pumps water through a hosepipe. Water passes through the pipe and leaves it with a velocity of $2 e

Question

An engine pumps water through a hosepipe. Water passes through the pipe and leaves it with a velocity of $2	ext{ m/s}$. The mass per unit length of water in the pipe is $100	ext{ kg m}^{-1}$. What is the power of the engine?

Accepted Answer

1. **Determine Mass Flow Rate:** The rate at which mass flows ($\frac{dm}{dt}$) is the product of mass per unit length ($\mu$) and the velocity ($v$).
   $$ \frac{dm}{dt} = \mu 	imes v = 100	ext{ kg m}^{-1} 	imes 2	ext{ m s}^{-1} = 200	ext{ kg s}^{-1} $$
2. **Calculate Power:** Power ($P$) is the rate at which kinetic energy is imparted to the water.
   $$ P = \frac{dK}{dt} = \frac{1}{2} \frac{dm}{dt} v^2 $$ [Class 11 Physics, Ch 6, Sec 6.10]
   Alternatively, substituting $\frac{dm}{dt} = \mu v$:
   $$ P = \frac{1}{2} (\mu v) v^2 = \frac{1}{2} \mu v^3 $$
3. **Substitute Values:**
   $$ P = \frac{1}{2} (100) (2)^3 = 50 	imes 8 = 400	ext{ W} $$

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