Answer to Question #273554 in Physics for Venus

Question

A 0.2kg is drag 0.5m upward with a uniform velocity along a plane inclined 30° with the horizontal by a force parallel to the incline. The coefficient of kinetic friction between the object and the plane is 0.25. How much work is done by the applied force? Show complete solution.

Answer 1

So, we have $h=0.5\ (m)$ , $m=0.2\ (kg)$ , $\alpha=30°$ , $\mu=0.25$ .

$W=\Delta E$

$W_F-W_{Fr}=\Delta E\to W_F=\Delta E+W_{Fr}=mgh+\mu mg\cos\alpha\cdot(h/\sin\alpha)=$

$=0.2\cdot9.8\cdot0.5+0.25\cdot0.2\cdot9.8\cdot\cos30°\cdot(0.5/\sin30°)=1.4\ (J)$ . Answer