A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line as \(v = k \sqrt{x}\), where k is a positive constant. The work done by all the forces acting on the particle, during the first t seconds is :
Answer & explanation
Correct answer: option 3
Given : \(v = k \sqrt{x} = \frac{dx}{dt}\)
Integrating : \(\int \frac{1}{\sqrt{x}} dx = k \int dt\)
\(\Rightarrow 2 \sqrt{x} = kt + C\)
Assuming : \(x(0) = 0 \text{; } \Rightarrow C = 0\)
\(2\sqrt{x} = kt \Rightarrow x = \frac{k^2 t^2}{4} \)
\(v = \frac{dx}{dt} = \frac{k^2 t}{2}\)
Thus, work done : \(W = \Delta KE\)
\(W = \frac{1}{2}mv^2 - 0\)
\(W = \frac{1}{2}m[\frac{k^2 t}{2}]^2\)
\(W = \frac{mk^4 t^2}{8}\)