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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Self Evaluation Test - Laws of Motion

  • question_answer
    A body of mass 5 kg under the action of constant force \[\vec{F}={{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}\] has velocity at \[t=0\text{ }s\] as \[\vec{v}=(6\hat{i}-2\hat{j})m/s\]and at \[t=\text{1}0\text{ }s\]gas\[\vec{v}=6\hat{j}\text{ m/s}\]. The force F is:

    A) \[\left( -3\hat{i}+4\hat{j} \right)\text{ N}\]           

    B) \[\left( -\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j} \right)\text{ N}\]

    C) \[\left( 3\hat{i}-4\hat{j} \right)\text{ N}\]  

    D) \[\left( \frac{3}{5}\hat{i}-\frac{4}{5}\hat{j} \right)\text{ N}\]

    Correct Answer: A

    Solution :

    [a] From question, Mass of body, m = 5 kg Velocity at \[t=0,\text{ }u=\text{ }\left( 6\hat{i}-2\hat{j} \right)\text{ }m/s\] Velocity at \[t=\text{ }10s,\text{ }v=+\text{ }6\hat{j}\text{ }m/s\] Force, F=? \[a=\frac{v-u}{t}=\frac{6\hat{j}-\left( 6\hat{i}-2\hat{j} \right)}{10}=\frac{-3\hat{i}+4\hat{j}}{5}\text{ m/}{{\text{s}}^{2}}\] Force, \[F=ma=5\times \frac{\left( -3\hat{i}+4\hat{j} \right)}{5}=\left( -3\hat{i}+4\hat{j} \right)\text{ N}\]


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