First slide

Power

Question

A particle of mass ‘M’ moved under a constant power Po. At same instant after the start, its speed is V and at a later instant, the speed is 2V; Neglecting friction, distance travelled by the particle as its speed increases from V to 2V is

Moderate

A

$\frac{12 M V^{3}}{5 P o}$
B

$\frac{8 P o V^{2}}{M}$
C

$\frac{9 M V^{3}}{5 P o}$
D

$\frac{7 M V^{3}}{3 P o}$

Solution

$p o w e r P = f o r c e . v e l o c i t y P_{0} = m a . v = m \frac{d v}{d t} v = m \frac{d v}{d t} \frac{d x}{d x} v = m \frac{d v}{d x} \frac{d x}{d t} v = m \frac{d v}{d x} v . v \frac{P_{0}}{M} d x = v^{2} d v i n t e g r a t i n g \int_{0}^{x} \frac{P_{0}}{M} d x = \int_{v}^{2 v} v^{2} d v \frac{P_{0}}{M} x = \frac{v^{3}}{3} \begin{array}{r} 2 v \\ v \end{array}\} x = \frac{M}{3 P o} [{(2 V)}^{3} - V^{3}] x = = \frac{7 M V^{3}}{3 P o}$

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