The initial potential energy
"PE_i=mgh"The final kinetic energy
"KE_f =\\frac{mv^2}{2}"The heat losses
"Q=PE_i-KE_f=mgh\\left(1-\\frac{v^2}{2gh}\\right)"Hence, the percentage of its initial potential energy which is deposited as heat
"\\frac{Q}{PE_i}=1-\\frac{v^2}{2gh}=1-\\frac{20^2}{2\\times 9.8\\times 30}=0.32=32\\%"
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