Kinematics

JEE Main level, numerical answer, medium

56 of 308

Question 56 of 308

A boy pushes a box of mass 2kg2kg with a force F⃗=(20i^+10j^)N\vec{F}=\left(20\hat{i}+10\hat{j}\right)N on a frictionless surface. If the box was initially at rest, then _______ mm is displacement along the xx-axis after 10s10s

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Answer: 500

F⃗=20i^+10j^\vec{F}=20\hat{i}+10\hat{j}
a⃗=F⃗m=20i^+10j^2⇒10i^+5j^\vec{a}=\dfrac{\vec{F}}{m}=\dfrac{20\hat{i}+10\hat{j}}{2}\Rightarrow 10\hat{i}+5\hat{j}
∴s⃗=12a⃗t2=12(10i^+5j^)×(10)2∴\vec{s}=\dfrac{1}{2}\vec{a}{t}^{2}=\dfrac{1}{2}\left(10\hat{i}+5\hat{j}\right)\times {\left(10\right)}^{2}
⇒50(10i^+5j^)m\Rightarrow 50\left(10\hat{i}+5\hat{j}\right)m
∴∴ Displacement along xx-axis
⇒50×10⇒500m\Rightarrow 50\times 10\Rightarrow 500m

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