Several forces act on the particle define the work w

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Several forces act on the particle. Define the work W done on the particle by a force F x for this trajectory, by W = Z x f x i F x dx Note: “W” stands for work, do not get confused with “weight” Units, in SI system: 1 Joule = 1 J = 1 N.m (same as K and U)
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Phys 2A - Mechanics Comments: Work can be defined for any one force acting on the particle, or for a sum of some forces, or even for the net force. One must simply be specific as to which it is. The time dependence of the trajectory, x=x(t), is irrelevant: Note that we may need to specify x(t) to compute work, but the result does not depend on it (so there may be many other forces acting on the particle dictating the actual time dependence in x=x(t)). Work depends on the trajectory itself, through: initial and final points the path followed: for example W = Z x f x i F x dx = Z t f t i F x dx dt dt = Z t f t i F x v x dt x i x f no the same as x i x f
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Phys 2A - Mechanics Examples: 1a. What is the work done by the gravitational force on a 20 kg box carried vertically up from y i =1.0m to y f =5.0m? W = Z y f y i F grav ,y dy = Z y f y i ( - mg ) dy = - mg ( y f - y i ) Notice that the box was moved arbitrarily, not thrown upwards! W = - (20)(9 . 8)(5 - 1) J = - 784 J ANS: 1b. What is the work done by the gravitational force on a 20 kg box carried vertically down from y i =5.0m to y f =1.0m? W = - (20)(9 . 8)(1 - 5) J = +784 J
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Phys 2A - Mechanics 2a. The same box is pushed against a 2.0 N/m spring to compress it from x=0.0cm to x=-5.0 cm. What is the work done by the spring? ANS: The choice of coordinate axis is implicit instatement of problem. So W = Z x f x i F spring ,x dx = Z x f x i ( - kx ) dx = - 1 2 k ( x 2 f - x 2 i ) W = - 1 2 (2 . 0)(( - 5 . 0) 2 - (0) 2 ) J = - 25 J Notice,again, that the box was moved arbitrarily, pushed by hand!
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