This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 7 CHAPTER 7  Work and Energy 1. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s 2 )(2.80 m) = 6.86 10 3 J . 2. The displacement is opposite to the direction of the retarding force, thus W = Fx cos 180 = (535 N)(1.25 10 3 m)( 1) = 6.69 10 5 J . 3. Because there is no acceleration, the contact force must have the same magnitude as the weight. The displacement in the direction of this force is the vertical displacement. Thus, W = F ? y = ( mg ) ? y = (65.0 kg)(9.80 m/s 2 )(20.0 m) = 1.27 10 4 J . 4. ( a ) Because there is no acceleration, the horizontal applied force must have the same magnitude as the friction force. Thus, W = F ? x = (230 N)(4.0 m) = 9.2 10 2 J . ( b ) Because there is no acceleration, the vertical applied force must have the same magnitude as the weight. Thus, W = F y = mg y = (1200 N)(4.0 m) = 4.8 10 3 J . 5. Because there is no acceleration, from the force diagram we see that F N = mg , and F = F fr = k mg . Thus, W = F x cos 0 = k mg x cos 0 = (0.50)(160 kg)(9.80 m/s 2 )(10.3 m)(1) = 8.1 10 3 J . 6. Because the speed is zero before the throw and when the rock reaches the highest point, the positive work of the throw and the (negative) work done by the (downward) weight must add to zero. Thus, W net = W throw + mgh cos 180 = 0, or h = W throw / mg cos 180 = (80.0 J)/(1.85 kg)(9.80 m/s 2 )(1) = 4.41 m . 7. 1 J = (1 kg m/s 2 )(1 m)(1000 g/kg)(100 cm/m) 2 = 1 10 7 g cm/s 2 = 1 10 7 erg . 1 J = (1 N m)(0.225 lb/N)(3.28 ft/m) = 0.738 ft lb . 8. The maximum amount of work that the hammer can do is the work that was done by the weight as the hammer fell: W max = mgh cos 0 = (2.0 kg)(9.80 m/s 2 )(0.50 m)(1) = 9.8 J . People add their own force to the hammer as it falls in order that additional work is done before the hammer hits the nail, and thus more work can be done on the nail. 9. If we assume the width of a cut is 0.30 m, and we cut the lawn parallel to the long side, the number of cuts required is N = (10.0 m)/(0.30m/cut) = 33 cuts, each 20 m long....
View
Full
Document
This note was uploaded on 03/16/2008 for the course PHYS phys230 taught by Professor Hadley during the Spring '08 term at A.T. Still University.
 Spring '08
 Hadley
 Energy, Force, Work

Click to edit the document details