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A body of mass 4 kg is moving with momentum of 8 kg.m.s-1. A force of 0.2 N acts on it in the direction of motion of the body for 10 s. The increase in kinetic energy is
- A10 J
- B8.5 J
- C4.5 J
- D4 J
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CCommunity
Momentum = Mass × Velocity
8 = 4 × u ⇒ u = 2 m/s
Force = The rate of change of momentum
$(F = \frac{{dP}}{{dt}} = \frac{{{m_2}{v_2} - {m_1}{v_1}}}{{dt}} = \frac{{m\left( {v - u} \right)}}{t})$
$(0.2 = \frac{{4\left( {v - 2} \right)}}{{10}} \Rightarrow 4\left( {v - 2} \right) = 2 \Rightarrow v = 2.5\;m/s)$
Increase in kinetic energy:
$(K.E = \frac{1}{2}m\left( {{v^2} - {u^2}} \right) = \frac{1}{2} \times 4 \times \left( {{{2.5}^2} - {2^2}} \right) = 4.5\;J)$
8 = 4 × u ⇒ u = 2 m/s
Force = The rate of change of momentum
$(F = \frac{{dP}}{{dt}} = \frac{{{m_2}{v_2} - {m_1}{v_1}}}{{dt}} = \frac{{m\left( {v - u} \right)}}{t})$
$(0.2 = \frac{{4\left( {v - 2} \right)}}{{10}} \Rightarrow 4\left( {v - 2} \right) = 2 \Rightarrow v = 2.5\;m/s)$
Increase in kinetic energy:
$(K.E = \frac{1}{2}m\left( {{v^2} - {u^2}} \right) = \frac{1}{2} \times 4 \times \left( {{{2.5}^2} - {2^2}} \right) = 4.5\;J)$