SATHEE: Work Energy And Power Question 314

Work Energy And Power Question 314

Question: A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from $v_{1}$ , to $v_{2}$ is (neglect friction)

Options:

A) $\frac{m}{3 P} (v_{2}^{3} - v_{1}^{3})$

B) $\frac{m}{3 P} (v_{2} - v_{1})$

C) $\frac{3 p}{m} (v_{2}^{2} - v_{1}^{2})$

D) $\frac{m}{3 P} (v_{2}^{2} - v_{1}^{2})$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $P = F v = m a v$

$\Rightarrow$ $a = \frac{P}{m v}$

$\Rightarrow$ $v \frac{d v}{d s} = \frac{P}{m v}$

$\Rightarrow$ $v^{2} d v = \frac{P}{m} d s$

$\Rightarrow$ $\frac{P}{m} \int_{0}^{s} d s = \int_{v_{1}}^{v_{2}} v^{2} d v$

$\Rightarrow$ $\frac{P}{m} s = \frac{1}{3} (v_{2}^{3} - v_{1}^{3})$

$\Rightarrow$ $s = \frac{m}{3 P} (v_{2}^{3} - v_{1}^{3})$

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Work Energy And Power Question 314

Question: A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from v1 , to v2 is (neglect friction)

Options:

Answer:

Solution:

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Question: A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from $v_{1}$ , to $v_{2}$ is (neglect friction)