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# 6.4 - 6.4 Work(Dated October 7 2011 Let x(t be the position...

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6.4: Work (Dated: October 7, 2011) Let x ( t ) be the position function of an object of mass m moving along a straight line as a function of time t . Then the force f on the object can be calculated from Newton’s second law: f ( t ) = m d 2 x dt 2 . In the SI metric system, the units for the mass, the displace- ment, the time, and the force are respectively kilograms (kg), meters (m), seconds (s), and newtons (N). Hence Newton’s second law says that a force of 1 N acting on a mass of 1 kg induces an acceleration of 1 m/s 2 . If the force is measured in pounds and the distance in feet, then the unit for the work is a foot-pount (ft-lb). The work W done by a constant force f to move an object a distance x is given by the product W = f x . In SI metric system, the work is measured in joules (J). In other words, 1 J = 1 N - m. The work W done by a continuous force f to move an ob- ject from x = a to x = b can be approximated by a Riemann sum and calculated as an integral. To see this, let’s divide the interval [ a , b ] into n subintervals of equal width Δ x = b - a n . Then the work W i done by the force f to move the object from x i - 1 to x i , where [ x i - 1 , x i ] is the i th subinterval, can be approximated by W i = f ( x * i ) Δ x , where x * i is a sample point in [ x i - 1 , x i ]. This approximation gets better as one decreases the width Δ x (by increasing the number of subintervals n ). This is so because f is contin- uous and thus doesn’t change much over sufficiently small interval. In consequence, the work W done by the force f in moving the object from x = a to x = b satisfies the condition W n X i = 1 W i = n X i = 1 f ( x * i ) Δ x . As one passes to the limit n → ∞ , one gets W = lim n →∞ n X i = 1 f ( x * i ) Δ x = Z b a f ( x ) dx . Example 1 (Exercise 6.4.3* in the tex) A particle is moved along the x-axis by a force that measures f ( x ) = 10 (1 + x ) 2 pounds at a point x feet from the origin. Find the work done W in moving the particle from the origin to a distance of 9 ft. Solution. W = Z 9 0 10 (1 + x ) 2 dx u = 1 + x = 10 Z 10 1 du u 2 = 9 ft-lb . Example 2 (Exercise 6.4.5 in the text) Refer to the text. Hint: Area under the graph. Ans: 180 J. Example 3 (Exercise 6.4.9** in the text) Hooke’s law states that the force f ( x ) required to maintain a spring stretched x units beyond its natural length is proportional to x: Hooke’s law: f ( x ) = kx .

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