- #1

Eclair_de_XII

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- 91

## Homework Statement

"A rope is used to pull a ##\frac{357}{100}kg## block at constant speed ##\frac{203}{50}m## across a horizontal floor. The force on the block from the rope is ##\frac{768}{100}N## and directed ##15°## above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and the floor?"

## Homework Equations

##W=ΔE_mec+ΔE_th=(ΔK+ΔU)+ΔE_th##

##ΔU=0##

##K=\frac{1}{2}mv^2##

##f_k=(mg)(sin\theta)##

## The Attempt at a Solution

(a) ##W=ΔK=F⋅Δx=(F)(cos\theta)(Δx)=(\frac{768}{100}N)(cos15°)(\frac{203}{50}m)=30.12J##.

(b) Then I use the relation ##W_{block-floor system}=-ΔE_{mec}+ΔE_{th}=0## to equate those two terms to each other, since the block releases mechanical energy, and gains thermal energy. Someone please correct me because I'm very sure that I am incorrect with this reasoning.

##ΔE_{th}=ΔE_{mec}=30.12J##

(c) Now I move onto finding the coefficient of static friction using the relation:

##ΔE_{th}=(f_k)(Δx)=(mg)(sin\theta)(μ_k)(Δx)##

##μ_k=\frac{ΔE_{th}}{(mg)(sin\theta)(Δx)}=\frac{(F)(cos\theta)(Δx)}{(mg)(sin\theta)(Δx)}=\frac{(F)(cot\theta)}{mg}=0.82##

I don't understand what it is that I'm doing wrong.