JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति JEE PYQ - Work Energy Power and Collision

  • question_answer
    A body of mass 5 kg under the action of constant force \[\overset{\to }{\mathop{F}}\,={{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}\]has velocity at\[t=0s\]as\[\overset{\to }{\mathop{v}}\,=\left( 6\hat{i}-2\hat{j} \right)m/s\]and at t = 10s as \[\overset{\to }{\mathop{v}}\,=+6\hat{j}\,m/s.\] The force \[\overset{\to }{\mathop{F}}\,\]is: [JEE ONLINE 11-04-2014]

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

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

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

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

    Correct Answer: A

    Solution :

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


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