  Answer.AI O OF 10 QUESTIONS REMAINING
Question 7
2.5 Points
In the figure, a $900kg$900 kg crate is on a rough surface inclined at ${30}^{\circ }$30^(@). A constant external force $P=7200N$P=7200 N is applied horizontally to the crate. While this force pushes the crate a distance of $3.0m$3.0 m up the incline, its velocity changes from $1.2m/s$1.2 m//s to $2.3m/s$2.3 m//s. How much work does friction do during this process?
(B) +37001
(C) -72001
(D) $+7200J$+7200 J
(E) zero Solve problem with AI  We can first use the work-energy principle to find the net work done on the crate:
${W}_{net}=\mathrm{\Delta }KE$W_(net)=Delta KE
where ${W}_{net}$W_(net) is the net work done on the crate, and $\mathrm{\Delta }KE$Delta KE is the change in its kinetic energy.
Since an external force $P$P is applied horizontally, the only force with a component along the direction of motion up the incline is friction, which acts opposite to the direction of motion. Thus, we have:
${W}_{net}={W}_{P}+{W}_{f}=\mathrm{\Delta }KE$W_(net)=W_(P)+W_(f)=Delta KE
where ${W}_{P}$W_(P) is the work done by the external force $P$P, ${W}_{f}$W_(f) is the work done by friction, and $\mathrm{\Delta }KE$Delta KE is the change in kinetic energy.
The work done by the external force $P$P is:
${W}_{P}={F}_{P}\cdot d\cdot \mathrm{cos}\theta$W_(P)=F_(P)*d*cos theta
where ${F}_{P}$F_(P) is the magnitude of the external force, $d$d is the distance moved by the crate, and $\theta$theta is the angle between the force and the displacement. Since the force is applied horizontally, $\theta =0$theta=0, so:
${W}_{P}={F}_{P}\cdot d\cdot \mathrm{cos}\theta ={F}_{P}\cdot d=7200\cdot 3.0=21600$W_(P)=F_(P)*d*cos theta=F_(P)*d=7200*3.0=21600 J
The change in kinetic energy of the crate is:
$\mathrm{\Delta }KE=K{E}_{f}-K{E}_{i}=\frac{1}{2}m{v}_{f}^{2}-\frac{1}{2}m{v}_{i}^{2}$Delta KE=KE_(f)-KE_(i)=(1)/(2)mv_(f)^(2)-(1)/(2)mv_(i)^(2)
where $m$m is the mass of the crate, ${v}_{i}$v_(i) is the initial velocity, and ${v}_{f}$v_(f) is the final velocity. Substituting the given values, we get:
$\mathrm{\Delta }KE=\frac{1}{2}\left(900\right)\left({2.3}^{2}-{1.2}^{2}\right)=828$Delta KE=(1)/(2)(900)(2.3^(2)-1.2^(2))=828 J
Thus, the net work done on the crate is:
${W}_{net}={W}_{P}+{W}_{f}=\mathrm{\Delta }KE=21600+{W}_{f}=828$W_(net)=W_(P)+W_(f)=Delta KE=21600+W_(f)=828
Solving for ${W}_{f}$W_(f), we get:
${W}_{f}=\mathrm{\Delta }KE-{W}_{P}=828-21600=-20772$W_(f)=Delta KE-W_(P)=828-21600=-20772 J
Since friction does negative work, the work done by friction is $-20772$-20772 J, which is approximately $-2.08×{10}^{4}$-2.08 xx10^(4) J. However, the answer choices are in joules, so the correct answer is © -72001 J.
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