A body of mass 5 kg under the action of constant force F→=Fxi^+Fyj^ . It has velocity at t=0 as v→=(6i^-2j^)m/s and at t=10 s as v→=+6j^m/s· The force F→ is

Acceleration, a=v−ut=6j^−(6i^−2j^)10=−3i^+4j^5 m/s2 Force, F=ma=5×(−3i^+4j^)5=(−3i^+4j^)N

A body of mass 5 kg under the action of constant force F→=Fxi^+Fyj^ . It has velocity at t=0 as v→=(6i^-2j^)m/s and at t=10 s as v→=+6j^m/s· The force F→ is

Acceleration, a=v−ut=6j^−(6i^−2j^)10=−3i^+4j^5 m/s2 Force, F=ma=5×(−3i^+4j^)5=(−3i^+4j^)N

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$\begin{array}{l} A body of mass 5 kg under the action of constant force \vec{F} {=F}_{x} \hat{i} {+F}_{y} \hat{j} . \\ It has velocity at t=0 as \vec{v} =(6 \hat{i} -2 \hat{j})m/s and at t=10 s as \vec{v} =+6 \hat{j} m/s· \\ The force \vec{F} is \end{array}$

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

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

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

$(- 3 \hat{j} + 4 \hat{j}) N$

answer is A.

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Detailed Solution

$\begin{array}{l} Acceleration, a = \frac{v - u}{t} = \frac{6 \hat{j} - (6 \hat{i} - 2 \hat{j})}{10} = \frac{- 3 \hat{i} + 4 \hat{j}}{5} m / s^{2} \\ Force, F = m a = 5 \times \frac{(- 3 \hat{i} + 4 \hat{j})}{5} = (- 3 \hat{i} + 4 \hat{j}) N \end{array}$