A body of mass $5 \mathrm{~kg}$ under the action of constant force $\vec{F}=F_x \hat{i}+F_y \hat{j}$ has…

A body of mass $5 \mathrm{~kg}$ under the action of constant force $\vec{F}=F_x \hat{i}+F_y \hat{j}$ has velocity at $\mathrm{t}=0 \mathrm{~s}$ as $\overrightarrow{\mathrm{v}}=(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s})$ and at $\mathrm{t}=10 \mathrm{~s}$ as $\overrightarrow{\mathrm{v}}=+6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}$. The force $\overrightarrow{\mathrm{F}}$ is:
  1. $(-3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}$
  2. $\left(-\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right) N$
  3. $(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \mathrm{N}$
  4. $\left(\frac{3}{5} \hat{\mathrm{i}}-\frac{4}{5} \hat{\mathrm{j}}\right) \mathrm{N}$

Solution

From question, Mass of body, $m=5 \mathrm{~kg}$ Velocity at $t=0$, $ u=(6 \hat{i}-2 \hat{j}) \mathrm{m} / \mathrm{s} $ Velocity at $t=10 \mathrm{~s}$, $ v=+6 \hat{j} \mathrm{~m} / \mathrm{s} $ Force, $F=$ ? 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} \mathrm{~m} / \mathrm{s}^2 $ Force, $F=m a$ $ =5 \times \frac{(-3 \hat{i}+4 \hat{j})}{5}=(-3 \hat{i}+4 \hat{j}) N $

Asked in: JEE Main 2014 (11 Apr Online)

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