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Chapter 6: Problem 88

In an adventure movie, a 62.5 -kg stunt woman falls \(8.10 \mathrm{m}\) and lands in a huge air bag. Her speed just before she hit the air bag was $10.5 \mathrm{m} / \mathrm{s} .$ (a) What is the total work done on the stunt woman during the fall? (b) How much work is done by gravity on the stunt woman? (c) How much work is done by air resistance on the stunt woman? (d) Estimate the magnitude of the average force of air resistance by assuming it is constant throughout the fall.

Short Answer

Expert verified
To find the total work done on the stunt woman during the fall, we will use the work-energy principle, and the given final speed of 10.5 m/s. Calculate the final kinetic energy using the formula: \(KE_f = \dfrac{1}{2} mv_f^2\) Subtract the initial kinetic energy (0, since the initial speed is 0) from the final kinetic energy: \(W_{total} = KE_f - KE_i\) Determine the total work done on the stunt woman during her fall.

Step by step solution

01

(Find the total work done on the stunt woman)

First, we will use the work-energy principle to find the total work done on the stunt woman during the fall. The work-energy principle states that the total work done on an object is equal to its change in kinetic energy: $$W_{total} = \Delta KE$$ We know her initial speed is 0 m/s (since she starts from rest), and her final speed is 10.5 m/s, as given in the problem. We can calculate her initial and final kinetic energies and find the difference: Initial kinetic energy \(KE_i = 0\) (because \(v_i = 0\)), Final kinetic energy \(KE_f = \dfrac{1}{2} mv_f^2\), $$W_{total} = KE_f - KE_i$$

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