Question #216722

A 10 kg block is pushed 20.0 m up the sloping surface of a plane inclined at an angle of 37o to the horizontal by a constant force F of 110.0 N acting parallel to the plane as shown in figure 2 below. The coefficient of kinetic friction between the block and plane is 0.25. Calculate work done by friction



1
Expert's answer
2021-07-14T09:55:42-0400

m=10kgm= 10 kg

h=20mh=20m

μ=0.25\mu= 0.25

α=37°\alpha=37 \degree

F=110N\vec F= 110N


we introduce the coordinate system:\text{we introduce the coordinate system:}

X-axis parallel to the slope of the inclined plane\text{X-axis parallel to the slope of the inclined plane}

Y-axis perpendicular to the X-axis\text{Y-axis perpendicular to the X-axis}

s the length of the slope of the inclined plane\text{s the length of the slope of the inclined plane}

s=hsinαs =\frac{h}{\sin \alpha}

According to Newton’s second law:\text{According to Newton's second law:}

ma=mg+N+F+Ffrm\vec a= m\vec g+\vec N+\vec F+\vec F_{fr}

projection of forces on the Y-axis:\text{projection of forces on the Y-axis:}

mgcosα+NFsinα=0-mg\cos\alpha +N -F\sin\alpha = 0

N=mgcosα+FsinαN = mg\cos\alpha +F\sin\alpha

Ffr=μNF_{fr}= \mu N

Ffr=μ(mgcosα+Fsinα)F_{fr}= \mu (mg\cos\alpha +F\sin\alpha )

Wfr=FfrsW_{fr}= F_{fr}*s

Wfr=μ(mgcosα+Fsinα)hsinαW_{fr}= \mu (mg\cos\alpha +F\sin\alpha )*\frac{h}{\sin \alpha}

Wfr=hμ(mgcotα+F)W_{fr}= h*\mu (mg\cot \alpha +F )

Wfr=200.25(109.8cot37°+110)=1200JW_{fr}= 20*0.25 *(10*9.8*\cot 37 \degree +110 )=1200J


Answer: Wfr=1200J\text{Answer: }W_{fr}=1200J



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