Question

Q11. A particle of mass 10 is moving with velocity of 10√x, where is displacement. The work done by net force during the displacement of particle form = 4 to = 9 is

Answer 1

AnswEr :

$\frak{Given} \begin{cases} \sf{Mass\ =\ 10\ kg} \\ \\ \sf{Velocity\ =\ 10 \sqrt{x}} \end{cases}$

SolutioN :

$\longrightarrow \tt{a\ =\ \dfrac{dv}{dt}} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{dv}{dt}\ \times\ \dfrac{dx}{dx}} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{dx}{dt}\ \times\ \dfrac{dv}{dx}} \\ \\ \\ \longrightarrow \tt{a\ =\ v \dfrac{dv}{dx}} \\ \\ \\ \longrightarrow \tt{a\ =\ 10 \sqrt{x} \dfrac{d(10 \sqrt{x})}{dx}} \\ \\ \\ \longrightarrow \tt{a\ =\ 10 \sqrt{x} \dfrac{10}{2 \sqrt{x}}} \\ \\ \\ \longrightarrow \tt{a\ =\ 10 \cancel{\sqrt{x}} \dfrac{5}{\cancel{\sqrt{x}}}} \\ \\ \\ \longrightarrow \tt{a\ =\ \dfrac{50}{1}} \\ \\ \\ \large \longrightarrow \underline{\boxed{\frak{a\ =\ 50\ ms^{-2}}}}$

__________________

$\longrightarrow \tt{W\ =\ F.dx} \\ \\ \\ \longrightarrow \tt{W\ =\ ma.dx} \\ \\ \\ \longrightarrow \tt{W\ =\ 10\ \times\ 50\ .dx} \\ \\ \\ \longrightarrow \tt{W\ =\ 10(50.dx)}$

Integrate both sides

$\displaystyle \longrightarrow \tt{\int W\ =\ 10 (\int_4 ^9 50.dx)} \\ \\ \\ \longrightarrow \tt{W\ =\ 10( \big[ 50x \big] ^9 _4)} \\ \\ \\ \longrightarrow \tt{W\ =\ 10(50 [9\ -\ 4]) } \\ \\ \\ \longrightarrow \tt{W\ =\ 10( 50[5])} \\ \\ \\ \large \longrightarrow \underline{\boxed{\frak{W\ =\ 2500\ J}}}$