A Body of mass 5 kg under the action of constant force F = Fx i + Fy j has velocity at t = 0 s as v = (6 i - 2 j) m/s and at t = 10 s as v = + 6 j m/s. The force F is: (a) ( - 3 i + 4 j) N (b) (-3/5i + 4/5j) N (c) (3 i - 4 j) N (d) (3/5i - 4/5j) N

Answer: (a). a = ()/(Δ t) = (6j - (6i - 2j))/(10) = (-6i + 8j)/(10) m/s² F = ma = 5 × (-0.6i + 0.8j) = -3i + 4j N So the answer is (a) (-3i + 4j) N.

A Body of mass 5 kg under the action of constant force F = Fx i + Fy… | JEE Main 2014 (11 Apr, Shift 1)

Q35JEE Main 2014 · 11 APR · SHIFT 1Easy

Inertia, First & Second Laws

A Body of mass 5 kg under the action of constant force $\vec{F}$ = Fx $\hat{i}$ + Fy $\hat{j}$ has velocity at t = 0 s 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:

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Answer: (a)

$\vec{a} = \frac{\Delta\vec{v}}{\Delta t} = \frac{6\hat{j} - (6\hat{i} - 2\hat{j})}{10} = \frac{-6\hat{i} + 8\hat{j}}{10}\text{ m/s}^2$
$\vec{F} = m\vec{a} = 5 \times (-0.6\hat{i} + 0.8\hat{j}) = -3\hat{i} + 4\hat{j}\text{ N}$
So the answer is (a) $(-3\hat{i} + 4\hat{j})$ N.

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Inertia, First & Second Laws question from JEE Main 2014 (11 Apr, Shift 1)

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