JEE Main 2021 Physics - Class 11 JEEM21Pheb3c Solution: A body of mass 2 kg moves under a force of N. It starts from…

\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

QJEEM21Pheb3c: JEE Main 2021 Physics - Class 11 JEEM21Pheb3c Solution: A body of mass 2 kg moves under a force of N. It starts from…