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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2021
  • Medium
  • Numerical

A boy pushes a box of mass \(2\, kg\) with a force \(\overrightarrow{ F }=(20 \hat{ i }+10 \hat{ j }) N\) on a frictionless surface. If the box was initially at rest, then ...........  \(m\) is displacement along the \(x-\)axis after \(10\, s\).

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Answer & Solution

Correct Answer

(B) \(500\)

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Detailed explanation

\(\overrightarrow{ F }=20 \hat{ i }+10 \hat{ j }\)

\(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=\frac{20 \hat{ i }+10 \hat{ j }}{2} \Rightarrow 10 \hat{ i }+5 \hat{ j }\)

\(\therefore \overrightarrow{ s }=\frac{1}{2} \overrightarrow{ a } t ^{2}=\frac{1}{2}(10 \hat{ i }+5 \hat{ j }) \times(10)^{2}\)

\(\Rightarrow 50(10 \hat{ i }+5 \hat{ j }) m\)

\(\therefore\) Displacement along \(x\) -axis

\(\Rightarrow 50 \times 10 \Rightarrow 500 m\)

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