A block of mass m = 1kg is placed on a frictionless horizontal surface. One end of a massless string of length 0.8 m is connected to the block. The string is straight and the other end of the string is moved vertically up with constant velocity $v_{0}$ . When the string makes an angle $θ = 30^{0}$ with the horizontal surface the block leaves the surface. Find the speed $v_{0}$ (in m/s) (given $g = 10 m / s^{2}$ )

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answer is 3.

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Detailed Solution

$v = v_{0} \tan θ a = v_{0} \sec^{2} θ \frac{d θ}{d t} y = l \sin θ v_{0} = l \cos θ \frac{d θ}{d t} a = \frac{{v_{0}}^{2}}{l} \sec^{3} θ T \cos θ = m a T \sin θ = m g a \tan θ = g \frac{{V_{0}}^{2}}{l} \sec^{3} θ \tan θ = g {V_{0}}^{2} = \frac{g l 3 \sqrt{3}}{8} \times \sqrt{3} V_{0} = 3$