Laws of Motion

JEE Main level, single correct, medium

51 of 210

Question 51 of 210

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:

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

From question,
Mass of body, m=5 kgm=5 \mathrm{~kg}
Velocity at t=0t=0,
u=(6i^−2j^)m/su=(6 \hat{i}-2 \hat{j}) \mathrm{m} / \mathrm{s}
Velocity at t=10 st=10 \mathrm{~s},
v=+6j^ m/sv=+6 \hat{j} \mathrm{~m} / \mathrm{s}
Force, F=F= ?
Acceleration, a=v−uta=\frac{v-u}{t}
=6j^−(6i^−2j^)10=−3i^+4j^5 m/s2=\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=(−3i^+4j^)N=5 \times \frac{(-3 \hat{i}+4 \hat{j})}{5}=(-3 \hat{i}+4 \hat{j}) N

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