JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 26)
A boy pushes a box of mass 2 kg with a force $$\overrightarrow F = \left( {20\widehat i + 10\widehat j} \right)N$$ on a frictionless surface. If the box was initially at rest, then ___________ m is displacement along the x-axis after 10s.
Answer
500
Explanation
$$\overrightarrow F = 20\widehat i + 10\widehat j$$
$$\overrightarrow a = {{\overrightarrow F } \over m} = {{20\widehat i + 10\widehat j} \over 2} = 10\widehat i + 5\widehat j$$
$$ \therefore $$ $$\overrightarrow s = {1 \over 2}\overrightarrow a {t^2} = {1 \over 2}\left( {10\widehat i + 5\widehat j} \right) \times {\left( {10} \right)^2}$$
$$ = 50\left( {10\widehat i + 5\widehat j} \right)m$$
$$ \therefore $$ Displacement along x-axis
= 50 $$\times$$ 10 = 500 m
$$\overrightarrow a = {{\overrightarrow F } \over m} = {{20\widehat i + 10\widehat j} \over 2} = 10\widehat i + 5\widehat j$$
$$ \therefore $$ $$\overrightarrow s = {1 \over 2}\overrightarrow a {t^2} = {1 \over 2}\left( {10\widehat i + 5\widehat j} \right) \times {\left( {10} \right)^2}$$
$$ = 50\left( {10\widehat i + 5\widehat j} \right)m$$
$$ \therefore $$ Displacement along x-axis
= 50 $$\times$$ 10 = 500 m
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