Question

A particle is moving along X axis under the influence of a force(F) which varies with position (X) as F is directly proportional to x power-1/5 the variation to power due to this force with X is
a) Xpower2/5 b)Xpower1/5 c)Xpower-1/5 d)X power1/10

Answer 1

Answer:x^1/10

Explanation:

F is directly proportional to x^-1/5.

So, a is directly proportional to x^-1/5.

Therefore,a=kx^-1/5;dv/dx=vdv/dx=kx^-1/4

Integration vdv=k integration x-1/5dx

Solve this you will get the answer!!

Answer 2

AnswEr :

Force acting on the particle varies with the displacement as :

$\sf \: F \propto \: {x}^{ - \frac{1}{5} }$

Thus,

$\implies \: \sf \: a \propto \: x {}^{ - \frac{1}{5} } \\ \\ \implies \: \sf \: a = \gamma x {}^{ - \frac{1}{5} } \\ \\ \implies \: \sf \: v \frac{dv}{dx} = \gamma {x}^{ - \frac{1}{5} } \\ \\ \implies \: \sf \: vdv = \gamma \: {x}^{ - \frac{1}{5} } dx$

Integrating on both sides,we get :

$\implies \: \sf \: \dfrac{ \: {v}^{2} }{2} = \gamma \times \dfrac{5x {}^{ \frac{4}{5} } }{4}$

Assigning the Proportionality Sign,

$\implies \: \sf \: v {}^{2} \propto {x}^{ \frac{4}{5} } \\ \\ \implies \: \sf \: v \propto {x}^{ \frac{4}{10} } \\ \\ \implies \: \sf \: v \propto \: {x}^{ \frac{2}{5} }$

We know that,

$\sf \: P = Fv \\ \\ \longrightarrow \: \sf \: P \propto \: ( {x}^{ - \frac{1}{5} } )( {x}^{ \frac{2}{5} } ) \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: P \propto \: x {}^{ \frac{1}{5} } }}$

Option (B) is correct