JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(10\, s\).
- A \(400\)
- B \(500\)
- C \(800\)
- D \(1200\)
Answer & Solution
Correct Answer
(B) \(500\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ F }=20 \hat{ i }+10 \hat{ j }\) \(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=\frac{20 \hat{ i }+10 \hat{ j }}{2} \Rightarrow 10 \hat{ i }+5 \hat{ j }\)…
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