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: 11. Picture the Problem : The wagon rolls horizontally but the force pulls upward at an angle. Strategy: Use equation 7-3 keeping in mind the angle between the force and the direction of motion. Solution: Use equation 7-3: ( 29 ( 29 o cos 16 N 10.0 m cos25 150 J 0.15 kJ W Fd θ = = = = Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force reduces the normal force a little. 25. Picture the Problem : The pine cone falls straight down for 16 m under the influence of gravity. Strategy: The work done by air resistance is the difference in kinetic energies between the air resistance and no air resistance cases. The work done by gravity is always W = mgh (see Example 7-2 and Conceptual Checkpoint 7-1). Solution: 1. (a) Find the difference in kinetic energies between the air resistance and no air resistance cases: ( 29 ( 29 ( 29 ( 29 2 1 f i f 2 2 2 1 2 0.140 kg 13 m/s 9.81 m/s 16 m W K K K mv mgh = ∆ =- =- =- = 2. (b) The work done by air resistance equals the average force of air resistance times the distance the pine cone falls. It is negative because the upward force is opposite to the downward distance traveled. ( 29 10 J so that 0.63 N up 16 m W W Fd F d- = - = - = - = Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion....
View Full Document
This note was uploaded on 01/05/2012 for the course CHEM 3321 taught by Professor Bean during the Spring '11 term at American Jewish University.
- Spring '11