Given:
Mass, m = 75 kg
Change in distance = 150
Acceleration, a1 = 35 kn/h
Acceleration. a2 = 45 km/h
Let's determine how much energy was dissipated by friction and air resistance during this descent.
To find how much energy was dissipated, apply the formula:
[tex](W\cdot Dg)+(W\cdot D_f)=\Delta KE[/tex]Rewrite the equation for Wdf:
[tex]WD_f=\Delta KE-WD_g[/tex]Where:
[tex]\begin{gathered} WD_g=75\times9.8\times150=110250\text{ J} \\ \\ \end{gathered}[/tex]Also:
[tex]\begin{gathered} \Delta KE=\frac{1}{2}\times75\times((\frac{45}{3.6})^2+(\frac{35}{3.6})^2) \\ \\ \Delta KE\text{ = }2314.81\text{ J} \end{gathered}[/tex]WHere Wdf is the energy dissipated, we have:
[tex]\begin{gathered} WD_f=\Delta KE-WDg \\ \\ WD_f=2314.81-110250=-107935\approx108\text{ KJ} \end{gathered}[/tex]ANSWER:
108 KJ
