196_PartUniversity Physics Solution

# 196_PartUniversity - Work and Kinetic Energy 6.49 6-13 W The work you do in lifting mass m a height h is mgh t SET UP 1 hp = 746 W EXECUTE(a The

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Work and Kinetic Energy 6-13 6.49. IDENTIFY: av W P t Δ = Δ . The work you do in lifting mass m a height h is mgh . SET UP: 1 hp 746 W = EXECUTE: (a) The number per minute would be the average power divided by the work ( mgh ) required to lift one box, 2 (0.50 hp) (746 W hp) 1.41 s, (30 kg) (9.80 m s ) (0.90 m) = or 84.6 min. (b) Similarly, 2 (100 W) 0.378 s, (30 kg) (9.80 m s ) (0.90 m) = or 22.7 min. EVALUATE: A 30-kg crate weighs about 66 lbs. It is not possible for a person to perform work at this rate. 6.50. IDENTIFY and SET UP: Use Eq.(6.15) to relate the power provided and the amount of work done against gravity in 16.0 s. The work done against gravity depends on the total weight which depends on the number of passengers. EXECUTE: Find the total mass that can be lifted: av , Wm g h P tt Δ == Δ so av Pt m gh = 4 av 746 W (40 hp) 2.984 10 W 1 hp P ⎛⎞ × ⎜⎟ ⎝⎠ 4 3 av 2 (2.984 10 W)(16.0 s) 2.436 10 kg (9.80 m/s )(20.0 m) m gh × = × This is the total mass of elevator plus passengers. The mass of the passengers is 33 2.436 10 kg 600 kg 1.836 10 kg. ×− = × The number of passengers is 3 1.836 10 kg 28.2. 65.0 kg × = 28 passengers can ride. EVALUATE: Typical elevator capacities are about half this, in order to have a margin of safety. 6.51. IDENTIFY: Calculate the gallons of gasoline consumed and from that the energy consumed. Find the time t Δ for the trip and use av W P t Δ = Δ , where W Δ is the energy consumed. SET UP: 200 km 124 mi = EXECUTE: (a) The gallons of gasoline consumed is 124 mi 4.13 gal 30 mi/gal = . The energy consumed is 99 (4.13 gal)(1.3 10 J/gal) 5.4 10 J ×= × . (b) The time for the trip is 124 mi 2.07 h 7450 s 60 mi/h . 9 5 av 7.2 10 W 720 kW 7450 s W P t Δ× = × = Δ . EVALUATE: The rate of energy consumption is 3 720 10 W 970 hp 746 W/hp × = . 6.52. IDENTIFY: Apply PF v = ! . F ! is the force F of water resistance. SET UP: 746 W = . 1 km/h 0.228 m/s = EXECUTE: 6 (0.70) (0.70) (280,000 hp)(746 W hp) 8.1 10 N. (65 km h) ((0.228 m/s) (1 km/h)) P F v = × EVALUATE: The power required depends on speed, because of the factor of v in v = ! and also because the resistive force increases with speed. 6.53. IDENTIFY: To lift the skiers, the rope must do positive work to counteract the negative work developed by the component of the gravitational force acting on the total number of skiers, rope sin FN m g α = . SET UP: rope vFv ! EXECUTE: ( ) rope rope cos vN m g v φ ⎡⎤ + ⎣⎦ . () ( ) 2 rope 1 m/s 50 riders 70.0 kg 9.80 m/s cos75.0 12.0 km/h 3.60 km/h P = . 4 rope 2.96 10 W 29.6 kW P = . EVALUATE: Some additional power would be needed to give the riders kinetic energy as they are accelerated from rest.

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6-14 Chapter 6 6.54. IDENTIFY: Relate power, work and time. SET UP: Work done in each stroke is WF s = and av PW t = . EXECUTE: 100 strokes per second means av 100 PF s t = with 1.00 s, 2 tF m g = = and 0.010 m. s = av 0.20 W. P = EVALUATE: For a 70 kg person to apply a force of twice his weight through a distance of 0.50 m for 100 times per second, the average power output would be 5 7.0 10 W × . This power output is very far beyond the capability of a person.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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196_PartUniversity - Work and Kinetic Energy 6.49 6-13 W The work you do in lifting mass m a height h is mgh t SET UP 1 hp = 746 W EXECUTE(a The

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