?Work-Energy Theorem The kinetic energy \(KE\) of an object of mass \(m\) moving with

Chapter 5, Problem 32

(choose chapter or problem)

Work-Energy Theorem The kinetic energy \(KE\) of an object of mass \(m\) moving with velocity \(v\) is defined as \(K E=\frac{1}{2} m v^{2}\) If a force \(f(x)\) acts on the object, moving it along the \(x-axis\) from \(x_{1}\) to \(x_{2}\), the Work-Energy Theorem states that the net work done is equal to the change in kinetic energy: \(\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}\), where \(v_{1}\) is the velocity at \(x_{1}\) and \(v_{2}\) is the velocity at \(x_{2}\).

How much work (in \(ft-lb\) ) is required to hurl a \(12 -lb\) bowling ball at \(20 \mathrm{mi} / \mathrm{h}\) ? (Note: Divide the weight in pounds by \(32 \mathrm{ft} / \mathrm{s}^{2}\), the acceleration due to gravity, to find the mass, measured in slugs.)

Equation Transcription:

Text Transcription:

KE

m

v

KE= 1/2 mv^2

f(x)

x-axis

x_1

x_2

ft-lb

12-lb

20 mi/h

32ft/s^2