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Chapter 6: Problem 29

Bruce stands on a bank beside a pond, grasps the end of a 20.0 -m-long rope attached to a nearby tree and swings out to drop into the water. If the rope starts at an angle of \(35.0^{\circ}\) with the vertical, what is Bruce's speed at the bottom of the swing?

Short Answer

Expert verified
Answer: Bruce's speed at the bottom of the swing is approximately \(12.52\,\mathrm{m/s}\).

Step by step solution

01

Calculate the initial height

To find the initial height (h) of Bruce above the bottom of the swing, we can use trigonometry, considering a right triangle formed by the initial position of the rope, its vertical component, and the horizontal component between Bruce and the vertical line passing through the tree attachment. The formula to find the height is h = L(1-cosθ), where L is the length of the rope and θ is the initial angle. \[h = 20.0\,\mathrm{m}\, (1 - \cos 35.0^{\circ})\] \[h \approx 7.987\,\mathrm{m}\]
02

Equate initial potential energy to final kinetic energy

As mentioned, the total mechanical energy is conserved. The initial potential energy (PE) is given by the equation PE = mgh, where m is the mass of Bruce, g is the gravitational acceleration (\(9.81\,\mathrm{m/s^2}\)), and h is the initial height. The final kinetic energy (KE) is given by the equation KE = 0.5mv^2, where m is the mass of Bruce and v is his final velocity at the bottom of the swing. The mass cancels out in both equations, so we have: \[gh = \frac{1}{2}v^2\]
03

Solve for v

Rearranging the equation to solve for v, we get: \[v = \sqrt{2gh}\] Plugging in the values for g and h, we can find Bruce's final velocity: \[v = \sqrt{2(9.81\,\mathrm{m/s^2})(7.987\,\mathrm{m})}\] \[v \approx 12.52\,\mathrm{m/s}\] Therefore, Bruce's speed at the bottom of the swing is approximately \(12.52\,\mathrm{m/s}\).

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Most popular questions from this chapter

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