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a 92.0 kg skydiver falls straight downward with an open parachute through a vertical height of 325 m. the skydivers velocity remains constant. what is the work done by the nonconservative force of air resistance, which is the only nonconservative force acting?
-2.93×10^5 j
answer is not obtainable, because insufficient information about the skydivers speed is given.
0 j
+2.93×10^5 j
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atter 2

Explanation:

Step1: Apply the work - energy theorem

The work - energy theorem is \(W_{nc}=\Delta K+\Delta U\). Since the skydiver's velocity remains constant, \(\Delta K = 0\) (because \(K=\frac{1}{2}mv^{2}\) and \(v\) is constant). The change in gravitational potential energy is \(\Delta U=U_{f}-U_{i}=-mgh\) (taking downwards as the direction of motion, \(h = 325m\), \(m = 92.0kg\), \(g=9.8m/s^{2}\)).

Step2: Calculate the work done by non - conservative force

\(W_{nc}=\Delta K+\Delta U\). Substituting \(\Delta K = 0\) and \(\Delta U=-mgh\) into the formula.

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Answer:

\(-2.93\times 10^{5}J\)