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

Chapter 5, Problem 33

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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 \(x_{2}\).

Suppose that when launching an \(800-kg\) roller coaster car an electromagnetic propulsion system exerts a force of \(\left(5.7 x^{2}+1.5 x\right)\) newtons on the car at a distance \(x\) meters along the track. Use Exercise \(31\) to find the speed of the car when it has traveled \(60\) meters.

Equation Transcription:

 

Text Transcription:

KE

m

v

KE= 1/2 mv^2

f(x)

x-axis

x_1

x_2

1/2 mv_2 ^2 - 1/2 mv_1^2

v_1

v_2

800-kg

5.7x^2+1.5x

31

60