Explanation:
Given, rate of decrease in kinetic energy,\(\dfrac{d K}{d t}=9.6 J^{-1}\)
When speed, \(v = 3\;m{s^{ - 1}},\) then force \(F=\) ?
Kinetic energy, \(K=\dfrac{1}{2} m v^{2}\)
Differentiating with respect to \(t\), we get
\(\frac{{dK}}{{dt}} = \frac{1}{2}\,\,\,m2v\,\,\frac{{dv}}{{dt}} = vm\frac{{dv}}{{dt}}\)
\( = vm\,\,\frac{{dp}}{{dt}}\)
\(\frac{{dK}}{{dt}} = vF\,\,\,\left[ {\because {\text{ force}},{\text{ }}F = m\frac{{dp}}{{dt}}} \right]\)
\(\therefore \quad F = \frac{1}{v} \cdot \frac{{dK}}{{dt}}\)
\( = \frac{1}{3} \times 9.6\)
\( = 3.2\;N\)