Answer to Question #216722 in Classical Mechanics for Miliza

Question

A 10 kg block is pushed 20.0 m up the sloping surface of a plane inclined at an angle of 37^o 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

Answer 1

$m= 10 kg$

$h=20m$

$\mu= 0.25$

$\alpha=37 \degree$

$\vec F= 110N$

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

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

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

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

$s =\frac{h}{\sin \alpha}$

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

$m\vec a= m\vec g+\vec N+\vec F+\vec F_{fr}$

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

$-mg\cos\alpha +N -F\sin\alpha = 0$

$N = mg\cos\alpha +F\sin\alpha$

$F_{fr}= \mu N$

$F_{fr}= \mu (mg\cos\alpha +F\sin\alpha )$

$W_{fr}= F_{fr}*s$

$W_{fr}= \mu (mg\cos\alpha +F\sin\alpha )*\frac{h}{\sin \alpha}$

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

$W_{fr}= 20*0.25 *(10*9.8*\cot 37 \degree +110 )=1200J$

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