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Work and Energy Problems

This problem involves calculating the average resistance force exerted on a 50 kg diver who drops from a 10 m high diving board and comes to rest 5 m below the water's surface. The average resistance force is determined to be 1470 N. Additionally, we analyze a circus performance where a monkey on a sled, weighing a combined 20 kg, is propelled up a 25° incline with an initial speed of 4 m/s. Considering kinetic friction, we find that the sled moves 1.4 m up the incline before coming to a stop.

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Work and Energy Problems

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  1. Work and Energy Problems

  2. A 50 kg diver steps off a 10 m high diving board and drops straight down into the water. If the diver comes to rest 5 m below the surface of the water, determine the average resistance force exerted on the diver by the water.

  3. #37,part 1 Work required to stop the diver W = Fdcosq W = (F)(5)(cos180) W = -5F 5 m

  4. #37, part 2 W = DTE Set PE=0 at bottom, KE= 0 at bottom ΔTE = 0 - PE at top of tower PEg = (50)(9.8)(15) = 7350 J ΔTE = -7350 W = -7350 -5F = -7350 F = 1470 N

  5. In a circus performance, a monkey on a sled is given an initial speed of 4 m/s up a 25º incline. The combined mass of the monkey and the sled is 20 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline does the sled move?

  6. #39 vi = 4 m/s m = 20 kg m = 0.20 d h = d sin 25 o 25

  7. #39, part 2 W = ΔTE = PEg - KE PEg = (20)(9.8)(dsin25) = 82.8 d KE = ½ (20)(4)2 = 160J WFf = Ff(d)(cos180) Ff = µmgcosθ WFf =(0.2)(20)(9.8)(cos25)(d)(-1)= -35.5d

  8. #39, part 3 W=ΔTE WFf =PEg – KE -35.5d =82.8d – 160 -118.3 d = -160 d = 1.4 m

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