If a body of mass 200 g falls from a height 200 m and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. (in J) of the body at the contact (g=10 m/s2)
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answer is 400.
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Detailed Solution
ΔU=mgh=0.2×10×200=400J∴Gain in K.E. = decrease in P.E. = 400 J.