A boy pushes a box of mass, $2 k g$ with a force, $F = 20 \vec{i} + 10 \vec{j} N$ on a frictionless surface. If the box was initially at rest, then what is displacement along the $x$ -axis after $10 s$ ?

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Solution

Step 1: Given data:
Mass of box, $m_{B} = 2 k g$
Force, $F = 20 \vec{i} + 10 \vec{j} N$
Time, $t = 10 s$
Step 2: Formula used:
Newton's second law: $F = m a$
Where, $a$ is the acceleration.
$m$ is the mass.
$F$ is the net force.
Step 3: Find the component of force in the horizontal and vertical directions:
Let the component of force in the horizontal direction be $F_{X}$ and the component in the vertical direction be $F_{Y}$
Therefore, from $F equals 20 i with rightwards arrow on top plus 10 j with rightwards arrow on top N$
$F_{X} = 20 N$ ,
$F_{Y} = 10 N$
Step 4: Find the acceleration in the horizontal and vertical directions:
Let the component of acceleration in the horizontal direction be $a_{X}$ and the component in the vertical direction be $a_{Y}$
$a_{X} = \frac{F_{X}}{m_{B}} = \frac{20}{2} = 10 m s^{- 2}$
$a_{Y} = \frac{F_{Y}}{m_{B}} = \frac{10}{2} = 5 m s^{- 2}$
Step 5: Find the displacement in the horizontal direction, $S_{X}$ :
Let the initial velocity of the box be $u_{X}$ which is zero
$S_{X} = u_{X} t + \frac{1}{2} a_{X} t^{2} S_{X} = 0 \times 10 s + \frac{1}{2} \times 10 m s^{- 2} \times {(10 s)}^{2} S_{X} = 500 m$
Hence, the displacement is the horizontal direction is $500 m$

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A boy pushes a box of mass, 2kg with a force, F=20i→+10j→N on a frictionless surface. If the box was initially at rest, then what is displacement along the x-axis after 10s?

A boy pushes a box of mass, $2 k g$ with a force, $F = 20 \vec{i} + 10 \vec{j} N$ on a frictionless surface. If the box was initially at rest, then what is displacement along the $x$ -axis after $10 s$ ?