115. (a) During one second, the decrease in potential energy is −=−=×=×ΔΔUmg y()(.55 106kg) 9.8m s (50 m)2.7 10 J29diwhere +yis upward and Δy= yf– yi.(b) The information relating mass to volume is not needed in the computation. By Eq. 8-40 (and the SI relation W = J/s), the result follows: P= (2.7
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