Staircase - escalator

An escalator moves downward at a speed of 0.6 m/s at an angle of 45° to the horizontal. A person weighing 80 kg walks upward on it at a speed of 0.9 m/s. Determine the distance covered by the person and the work done by him before he reaches a height of 20 m.

Final Answer:

s =  28.2843 m
W =  16000 J

Step-by-step explanation:

$$\begin{gathered} h=20\text{m} \\ \alpha =45{}^{\circ } \\ {v}_1=0.6\text{m/s} \\ {v}_2=0.9\text{m/s} \\ v=v_2-v_1=0.9-0.6=\frac{3}{10}=0.3\text{m/s} \\ \sin \alpha =v_y:v \\ {v}_y=v\cdot \sin \alpha =v\cdot \sin 45°=0.3\cdot \sin 45°=0.3\cdot 0.707107=0.21213\text{m/s} \\ {v}_x=v/\sqrt{2}=0.3/\sqrt{2}\doteq 0.2121\text{m/s} \\ {v}_y=v/\sqrt{2} \\ {v}_y=h/t \\ t=h/v_y=20/0.2121\doteq 94.2809\text{s} \\ s=v\cdot t=0.3\cdot 94.2809=20\sqrt{2}=28.2843\text{m} \end{gathered}$$

$$\begin{gathered} m=80\text{kg} \\ g=10{\text{m/s}}^2 \\ W=m\cdot g\cdot h=80\cdot 10\cdot 20=16000\text{J} \end{gathered}$$

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