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