- Home
- Standard 11
- Physics
4-1.Newton's Laws of Motion
medium
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
$1200$
(JEE MAIN-2021)
Solution
$\overrightarrow{ F }=20 \hat{ i }+10 \hat{ j }$
$\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }=\frac{20 \hat{ i }+10 \hat{ j }}{2} \Rightarrow 10 \hat{ i }+5 \hat{ j }$
$\therefore \overrightarrow{ s }=\frac{1}{2} \overrightarrow{ a } t ^{2}=\frac{1}{2}(10 \hat{ i }+5 \hat{ j }) \times(10)^{2}$
$\Rightarrow 50(10 \hat{ i }+5 \hat{ j }) m$
$\therefore$ Displacement along $x$ -axis
$\Rightarrow 50 \times 10 \Rightarrow 500 m$
Standard 11
Physics
