6.6 Notes (Work Part 1)

6.6 Notes (Work Part 1) - Hookes Law: The force F required...

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Work Part 1- Section 6.6 Constant Force: The amount of work done by a constant force F is W = Fd where d is the distance an object is moved by a force F . Units: Force: lb or N= kg · m/s 2 (Newtons) Work: ft-lb, in-lb or J=N · m (Joules) Example 1: A 150 lb man is climbing a 20 ft pole. Calculate the work done by the man to reach the top. Example 2: A 750 ft cable weighing 6 lb/ft is connected to and elevator weighing 1500 lb. Find the work done to lift the elevator 500 ft. Variable Force: The work done by a force F ( x ) as an object is moved from a to b is W = Z b a F ( x ) dx
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Back to Example 2: What work do you need to show for these types of problems? 1. 2. 3. 4. Example 3: A mountain climber is about to haul up a 50m length of hanging rope. How much work will it take if the rope weighs .6 N/m ?
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Unformatted text preview: Hookes Law: The force F required to compress or stretch a spring is given by F = kx where k is the spring constant and x is the amount the spring is stretch or compressed from its natural length. Example 4: A spring of natural length 12 is stretched to 15 under a weight of 10 lb. (a) Find the work done (units of in-lbs) to stretch it from 12 to 13. (b) Find the work done to stretch it from 15 to 18. Example 5: It takes 100 in-lb of work to stretch a spring from its natural length of 10 to 20. (a) Determine the spring constant k . (b) Determine the work done to stretch the spring from 12 to 14....
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6.6 Notes (Work Part 1) - Hookes Law: The force F required...

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