Verified Solution • 2025–26 Session moderate

QJEEM21Ph7758: JEE Main 2021 Physics - Class 11 JEEM21Ph7758 Solution: A boy pushes a box of mass 2 kg with a force on a frictionless…

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

Step-by-Step Verified Solution

F→=20i^+10j^\overrightarrow F = 20\widehat i + 10\widehat j

a→=F→m=20i^+10j^2=10i^+5j^\overrightarrow a = {{\overrightarrow F } \over m} = {{20\widehat i + 10\widehat j} \over 2} = 10\widehat i + 5\widehat j

∴\therefore s→=12a→t2=12(10i^+5j^)×(10)2\overrightarrow s = {1 \over 2}\overrightarrow a {t^2} = {1 \over 2}\left( {10\widehat i + 5\widehat j} \right) \times {\left( {10} \right)^2}

=50(10i^+5j^)m= 50\left( {10\widehat i + 5\widehat j} \right)m

∴\therefore Displacement along x-axis

= 50 ×\times 10 = 500 m

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