Physics · 2014
JEE Main · 11 April 2014, Shift 1 · Q3
A body of mass 5 kg under the action of constant force F = F_x i + F_y 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 /…
A body of mass $\displaystyle 5$ kg under the action of constant force $\displaystyle \overrightarrow{\mathrm{F}}=\mathrm{F}_x \hat{i}+\mathrm{F}_y \hat{j}$ has velocity at $\displaystyle \mathrm{t}=0 \mathrm{~s}$ as $\displaystyle \vec{v}=(6 \hat{i}-2 \hat{j}) \mathrm{m} / \mathrm{s}$ and at $\displaystyle \mathrm{t}=10 \mathrm{~s}$ as $\displaystyle \vec{v}=+6 \hat{j} \mathrm{~m} / \mathrm{s}$. The force $\displaystyle \overrightarrow{\mathrm{F}}$ is :
Official answer
(1)
$\displaystyle (-3 \hat{i}+4 \hat{j}) \mathrm{N}$
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JEE Main 2014 Physics question, with the answer from NTA’s final answer key. Where our answers come from.