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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 = 0s as \vec{v}=\left ( 6\hat{i}-2\hat{j} \right )m/s and at t = 10 s as \vec{v}=+6\hat{j}\: m/s. The force \vec{F} is :

  • Option 1)

    \left ( -3\hat{i}+4\hat{j} \right )N

  • Option 2)

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

  • Option 3)

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

  • Option 4)

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

 

Answers (1)

  In x direction \Delta p_x = m (v_{xf}-v_{xi})= 5 (0-6)= -30 Kg m /s

           

In y direction \Delta p_y = m (v_{yf}-v_{yi})= 5 (6-(-2))= 40 Kg m /s

F_x \Delta t= \Delta p_x \Rightarrow F_x = -3 N

F_y \Delta t= \Delta p_y \Rightarrow F_y = 4 N

F = -3 \hat i +4 \hat j

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Safeer PP

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