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A body of mass 0.3 kg is taken up an inclined plane of length 10 m and height 5 m and then allowed to slide down to bottom again. Find kinetic energy at the end of the trip. |
Answer» Solution : `sintheta=(5)/(10)=(1)/(2)=sin.(PI)/(6)` so, `THETA=(pi)/(6)` P.E at top = `0.3xx9.8xx5=14.7J` `{:("K.E at the end"),("of TRIP"):}}={{:("P.E at top"),("- Work done"),("against friction"):}` `K.E=14.7-7.6` `=7.1J` |
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