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Name: The muzzle velocity of a typical 500 g spud is 25 m/s . The force give
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Description: The muzzle velocity of a typical 500 g spud is 25 m/s . The force given by the spud is K/(x+.09) where x is the distance of the barrel in meters. If the length of the barrel is 75 cm, what is the constant K? What is the force on the spud?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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The muzzle velocity of a typical 500 g spud is 25 m/s . The force given by the spud is K/(x+.09) where x is the distance of the barrel in meters. If the length of the barrel is 75 cm, what is the constant K? What is the force on the spud?

Expert Solution

Step 1

Work done on spud = change in kinetic energy

$\int \vec{F} . \vec{d S}$ =K_Final-K_Initial

$\int F d x \cos 0 = \frac{1}{2} m v_{F l i n a l}^{2} - \frac{1}{2} m v_{I n i t i a l}^{2}$ Here $F = \frac{K}{x + 0.09}$ , V_Final=25 m/s , V_Initial=0 , m=500g=0.5 Kg

$\int_{0}^{0.75} \frac{K}{x + 0.09} d x = \frac{1}{2} \times 0.5 \times {(25)}^{2} - 0$

${[K \log_{e} (x + 0.09)]}_{0}^{0.75} = 156.25$

K(log_e(0.75+0.09)-log_e(0+0.09))=156.25

Using $\log_{e} m - \log_{e} n = \log_{e} \frac{m}{n}$

$K \log_{e} \frac{0.84}{0.09} = 156.25$

K×2.23=156.25

K= 70.067

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