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JEE Main 201411 AprilOnline
Exam
JEE Main
Year
2014
Date
11 April
Mode
Online
Subject
Physics
Chapter
Laws of Motion

JEE Main 2014 · 11 April · Online · Physics · Laws of Motion

A body of mass 5 ~kg under the action of constant force F =F_x i...

A body of mass 5 kg5 \mathrm{~kg} under the action of constant force F⃗=Fxi^+Fyj^\vec{F}=F_x \hat{i}+F_y \hat{j} has velocity at t=0 s\mathrm{t}=0 \mathrm{~s} as v→=(6i^−2j^m/s)\overrightarrow{\mathrm{v}}=(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}) and at t=10 s\mathrm{t}=10 \mathrm{~s} as v→=+6j^m/s\overrightarrow{\mathrm{v}}=+6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}. The force F→\overrightarrow{\mathrm{F}} is:

Options

  1. A(−3i^+4j^)N(-3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}
  2. B(−35i^+45j^)N\left(-\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right) N
  3. C(3i^−4j^)N(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \mathrm{N}
  4. D(35i^−45j^)N\left(\frac{3}{5} \hat{\mathrm{i}}-\frac{4}{5} \hat{\mathrm{j}}\right) \mathrm{N}

Correct answer

A. (−3i^+4j^)N(-3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}

Step-by-step solution

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