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An archer pulls her bowstring back \(0.400 \mathrm{~m}\) by exerting a force that increases uniformly from zero to \(230 \mathrm{~N}\). (a) What is the equivalent spring constant of the bow? (b) How much work is done in pulling the bow?

Short Answer

Expert verified
The equivalent spring constant of the bow is \(575N/m\) and the work done in pulling the bow is \(46J\)

Step by step solution

01

Identify Knowns and Unknowns

In this exercise, the final force applied (\(F\)) is \(230N\), and the displacement (\(x\)) is \(0.400m\). We need to find the equivalent spring constant (\(k\)) of the bow and the total work done in pulling the bow.
02

Calculate Equivalent Spring Constant with Hooke's Law

Using Hooke's Law (\(F=kx\)), we can solve for the spring constant (\(k\)) by rearranging the formula to \(k=F/x\). Substituting \(F = 230N\) and \(x = 0.400m\) into \(k=F/x\) we get \(k = 230N/0.400m = 575 N/m\). Therefore, the equivalent spring constant of the bow is \(575N/m\).
03

Calculate the Work Done in Pulling the Bow

The total work done (W) in pulling the bow can be calculated using the formula \(W = 1/2 kx^2\). Substituting \(k = 575 N/m\) and \(x = 0.400m\) into \(W = 1/2 kx^2\), we get \(W = 1/2 (575N/m) * (0.400m)^2 = 46J\). Therefore, 46J of work is done in pulling the bow.

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