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At the moment $\mathrm{t}=0$, a time dependent force $\mathrm{F}=$ at (where a is constant equal to $1 \mathrm{Ns}^{-1}$ ) is applied to a body of mass $1 \mathrm{~kg}$ resting on a smooth horizontal plane as shown in the figure. If the direction of this force makes an angle $45^{\circ}$ with the horizontal, then the velocity of the body at the moment it leaves the plane is
(acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
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(acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
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Verified Answer
The correct answer is:
$50 \sqrt{2} \mathrm{~ms}^{-1}$
When body left the surface, $\mathrm{N}=0$
So, $m g=$ at $\sin 45^{\circ}$
$\mathrm{t}=\frac{\mathrm{mg}}{\mathrm{a} \sin 45^{\circ}}=\frac{10}{\frac{1.1}{\sqrt{2}}}=10 \sqrt{2} \mathrm{sec}$
Now, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \Rightarrow \mathrm{dv}=\mathrm{adt}$
$\begin{aligned} & \Rightarrow \int_0^{\mathrm{v}} \mathrm{dv}=\int \mathrm{at} \cos 45^{\circ} \mathrm{dt} \Rightarrow \mathrm{v}=\frac{\mathrm{a}}{\sqrt{2}}\left[\frac{\mathrm{t}^2}{2}\right]_0^{10 \sqrt{2}} \\ & \Rightarrow \mathrm{v}=\frac{\mathrm{a}}{\sqrt{2}} \times \frac{1}{2}[200-0] \\ & \Rightarrow \mathrm{v}=50 \sqrt{2} \mathrm{~m} / \mathrm{s}\end{aligned}$
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