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The kinetic energy $ KE $ of an object of mass m moving with velocity $ v $ is defined as $ KE = \frac{1}{2} mv^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} mv^2_2 - \frac{1}{2} mv^2_1$ , where $ v_1 $ is the velocity at $ x_1 $ and $ v_2 $ is the velocity at $ x_2 $.

(a) 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

$$ W = \int_{x_1}^{x_2} f(x) dx = \int_{t_1}^{t_2} f(s(t)) v(t) dt $$

The use Newton's Second Law of Motion (force = mass $ \times $ acceleration) and the substitution $ u = v(t) $ to evaluate the integral.

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

(a) $\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}$

(b) $161 . \overline{3} \mathrm{ft}- \mathrm{l

b}$

Applications of Integration

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University of Illinois at Urbana-Champaign

Applications of Integration