Percentage of Block Energy Dissipated

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SUMMARY

The discussion focuses on calculating the percentage of energy dissipated by a block due to air resistance during a vertical fall. The block's initial speed is 6 m/s, and it increases to 8 m/s over a distance of 3 m. The calculations show that 33% of the block's energy is dissipated, derived from the equations of kinetic and potential energy. The final result indicates that the dissipated energy is accurately represented as 0.33 or 33% of the initial energy.

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Masrat_A

Homework Statement


During a vertical fall, a block's speed increases from 6 m/s to 8m/s over a 3 m distance. What percentage of the block's energy is dissipated due to air resistance over this distance?

Homework Equations


##E_f = E_i##

The Attempt at a Solution


##E_f = DE_i##
##KE_f + PE_f = D(KE_i + PE_i)##
##1/2MV_f^2 + 0 = D(1/2MV_i^2 + Mgy)##
##V_f^2 = D(V_i^2 + 2gy)##
##64 = D(36 + 60)##
##D = 0.67##

##1 - 0.67 = 0.33##
##0.33 * 100 = 33##

I wasn't so sure on what to label dissipated energy, so I went with D. Does any of this seem reasonable?
 
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It's all very reasonable.
 

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