Practicing Success
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 : |
\(\frac{mk^4 t^2}{16}\) \(\frac{mk^4 t^2}{4}\) \(\frac{mk^4 t^2}{8}\) \(\frac{mk^4}{t^2}\) |
\(\frac{mk^4 t^2}{8}\) |
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 : \(W = \Delta KE\) \(W = \frac{1}{2}mv^2 - 0\) \(W = \frac{1}{2}m[\frac{k^2 t}{2}]^2\) \(W = \frac{mk^4 t^2}{8}\) |