A force acts on a block as shown in figure. Find time when block loses contact with surface.
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$\mathrm{N}=\mathrm{F} \sin 37^{\circ}-\mathrm{mg}$

as per given condition

When $\mathrm{N}=0 \quad \mathrm{F} \sin 37^{\circ}-\mathrm{mg}=0$

$\Rightarrow F \sin 37^{\circ}=\mathrm{mg}$

$\Rightarrow 10 t \times \frac{3}{5}=10 \times 10 \Rightarrow t=\frac{50}{3} \mathrm{s}$

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