JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 21)
A body of mass 2 kg moves under a force of $$\left( {2\widehat i + 3\widehat j + 5\widehat k} \right)$$N. It starts from rest and was at the origin initially. After 4s, its new coordinates are (8, b, 20). The value of b is _____________. (Round off to the Nearest Integer)
Answer
12
Explanation
$$\overrightarrow F = (2\widehat i + 3\widehat j + 5\widehat k)N$$
time = 4 sec
As body start from rest therefore
position vector initially $$\overrightarrow {{r_i}} = (0\widehat i + 0\widehat j + 0\widehat k)$$ &
u (initial velocity) = 0
given, $${r_f} = (x\widehat i + y\widehat j + z\widehat k)$$
Now, from second equation of motion
$$\overrightarrow s = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$$
$${r_f} - {r_i} = {1 \over 2} \times \left( {{{2\widehat i + 3\widehat j + 5\widehat k} \over 2}} \right) \times {(4)^2}$$
$$ \Rightarrow $$ $$(x\widehat i + y\widehat j + z\widehat k) - (0\widehat i + 0\widehat j + 0\widehat k) = 8\widehat i + 12\widehat j + 20\widehat k$$
$$ \Rightarrow $$ $$x\widehat i + y\widehat j + z\widehat k = 8\widehat i + 12\widehat j + 20\widehat k$$
$$ \therefore $$ The value of b = 12
time = 4 sec
As body start from rest therefore
position vector initially $$\overrightarrow {{r_i}} = (0\widehat i + 0\widehat j + 0\widehat k)$$ &
u (initial velocity) = 0
given, $${r_f} = (x\widehat i + y\widehat j + z\widehat k)$$
Now, from second equation of motion
$$\overrightarrow s = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$$
$${r_f} - {r_i} = {1 \over 2} \times \left( {{{2\widehat i + 3\widehat j + 5\widehat k} \over 2}} \right) \times {(4)^2}$$
$$ \Rightarrow $$ $$(x\widehat i + y\widehat j + z\widehat k) - (0\widehat i + 0\widehat j + 0\widehat k) = 8\widehat i + 12\widehat j + 20\widehat k$$
$$ \Rightarrow $$ $$x\widehat i + y\widehat j + z\widehat k = 8\widehat i + 12\widehat j + 20\widehat k$$
$$ \therefore $$ The value of b = 12
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