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 \(10\, s\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(500\)
Step-by-step Solution
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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