Work, Power and EnergyHardBloom L3
Question
A force exerts an impulse I on a particle changing its speed from u to 2u. The applied force and the initial velocity are oppositely directed along the same line. The work done by the force is
Options
A.$\frac{3}{2}$ I u
B.$\frac{1}{2}$I u
C.I u
D.2I u
Solution
Sol. 
⇒ mu – I = –2mu ⇒ I = 3mu
ω = ∆KE= $\frac{1}{2}m(2u)^{2} - \frac{1}{2}mu^{2} = \frac{3}{2}mu^{2} = \left( \frac{3}{2}mu \right).u$
$\omega = \frac{I}{2}.u$
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