?Work-Energy Theorem The kinetic energy KE of an object of mass moving with velocity

Chapter 5, Problem 31

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Work-Energy Theorem The kinetic energy KE of an object of mass  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 \(x_{2}\).

Let \(x=s(t)\) be the position function of the object at time \(t\) and \(v(t)\), \(a(t)\) the velocity and acceleration functions. Prove the Work-Energy Theorem by first using the Substitution Rule for Definite Integrals \((5.5.6)\) to show that

\(\mathrm{W}=\int_{x_{1}}^{x_{2}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{t_{1}}^{t_{2}} \mathrm{f}(\mathrm{s}(\mathrm{t})) \mathrm{v}(\mathrm{t}) \mathrm{dt}\)

Then use Newton's Second Law of Motion (force \(=\) mass \(\mathrm {x}\) acceleration) and the substitution \(u=v(t)\) to evaluate the integral.

Equation Transcription:

v

KE =  mv2

f(x)

x1

x2

v1

v2

x = s(t)

t

v(t)

a(t)

W = ∫  f(x) dx = ∫  f(s(t)) v(t) dt

u = v(t)

Text Transcription:

v

KE = 1/2 mv^2

f(x)

x_1

x_2

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

v_1

v_2

x = s(t)

t

v(t)

a(t)

W = integral_x_1 ^x_2 f(x) dx = integral_t_1 ^t_2 f(s(t)) v(t) dt

u = v(t)