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}{3P}(v_2^{3}-v_1^{3}) $
B) $ \frac{m}{3P}(v_2-v_1) $
C) $ \frac{3p}{m}(v_2^{2}-v_1^{2}) $
D) $ \frac{m}{3P}(v_2^{2}-v_1^{2}) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ P=Fv=mav $
$ \Rightarrow $ $ a=\frac{P}{mv} $
$ \Rightarrow $ $ v\frac{dv}{ds}=\frac{P}{mv} $
$ \Rightarrow $ $ v^{2}dv=\frac{P}{m}ds $
$ \Rightarrow $ $ \frac{P}{m}\int _{0}^{s}{ds}=\int _{v_1}^{v_2}{v^{2}dv} $
$ \Rightarrow $ $ \frac{P}{m}s=\frac{1}{3}(v_2^{3}-v_1^{3}) $
$ \Rightarrow $ $ s=\frac{m}{3P}(v_2^{3}-v_1^{3}) $
