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Chapter 7: Problem 21

A 120 -g arrow is shot vertically from a bow whose effective spring constant is \(430 \mathrm{N} / \mathrm{m}\). If the bow is drawn \(71 \mathrm{cm}\) before shooting, to what height does the arrow rise?

Short Answer

Expert verified
The arrow rises to approximately 105 meters

Step by step solution

01

Convert Measurements to Basic SI Units

The mass of the arrow provided is in grams and needs to be converted to kilograms (multiply by \(1e-3\)). The drawn length of the bow is provided in cm and should be converted to meters (multiply by \(0.01\)). Mass \(m = 120g = 0.12kg\), and draw length of the bow \(x = 71cm = 0.71m\).
02

Set Up the Equation

Since the energy initially stored in the bow is transformed into the gravitational potential energy at maximum height, we can equate these two forms of energy. Thus the equation becomes: \(0.5kx^2 = mgh\), where \(k\) is the spring constant, \(x\) is the displacement, \(m\) is the mass, \(g\) is the acceleration due to gravity (9.8 m/s虏 approx), and \(h\) is the height.
03

Solve for the Maximum Height (h)

Re-arrange the equation to solve for \(h\): \(h = \(\frac{{0.5kx^2}}{{mg}}\). Insert the known values: \(h = \(\frac{{0.5 * 430 * (0.71)^2}}{{0.12 * 9.81}}\).

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