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Q.

A small block of mass 0⋅1 kg is pressed against a horizontal spring fixed at one end to compress the spring through 5⋅0 cm as shown in figure.The spring constant is 100 N/m. When released the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring ?

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a

at a horizontal distance of 1 m from free end of the spring

b

at a horizontal distance of 2 m from free end of the spring

c

vertically below the edge on which the mass is resting

d

at a horizontal distance of 2m from free end of the spring

answer is A.

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Detailed Solution

The energy stored in spring will be given to mass as kinetic energy. Hence12kx2=12mv212×100×51002=12×0⋅1×v2Solving, we get            V=(5/2)Now        y=2=0+12gt2or       t=(4/g)=(2/5)Further,        X=Vt=52×25=1m
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