JEE Main 2014 · 11 Apr · Shift 1 · Physics Q3

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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 $t=0$ s 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 :

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

\left(-\dfrac35\hat i+\dfrac45\hat j\right)

\left(3\hat i-4\hat j\right)

\left(\dfrac35\hat i-\dfrac45\hat j\right)

Answer and worked solution

Official answer: A

$\vec a=\frac{\Delta\vec v}{\Delta t}=\frac{(6\hat j)-(6\hat i-2\hat j)}{10}=-0.6\hat i+0.8\hat j,\qquad\vec F=m\vec a=\left(-3\hat i+4\hat j\right)\text{ N}.$

Topics:Newton's laws Vectors

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 3