211_PartUniversity Physics Solution

# 211_PartUniversity Physics Solution - 6-28 Chapter 6...

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6-28 Chapter 6 EXECUTE: (a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen consumption rate appears to be about 3 12 cm /kg min, and so the total energy is 33 5 (12 cm /kg min) (70 kg) (12 min) (20 J/cm ) 2.0 10 J. ⋅= × (b) The run will take 6 min. Using an estimation of the rate from the graph of about 3 33 cm /kg min gives an energy consumption of about 5 2.8 10 J. × (c) The run takes 4 min, and with an estimated rate of about 3 50 cm /kg min, the energy used is about 5 × (d) Walking is the most efficient way to go. In general, the point where the slope of the line from the origin to the point on the graph is the smallest is the most efficient speed; about 5 km/h. EVALUATE: In an exercise program, for a fixed distance, running burns more energy than walking. 6.104. IDENTIFY: Write equations similar to (6.11) for each component. Eq.(6.12) will now involve the sum of three integrals, one for each component. SET UP: 2222 x yz vvvv =++ EXECUTE: From , , x xy y mF m aF m a === Fa " " and . z z Fm a = The generalization of Eq. (6.11) is then . y x z xx yy zz dv dv dv av dx dy dz == = The total work is then 222 2 2 2 111 1 1 1 (,,) tot xyz x y z y xz x y z x y z dv dv dv W F dx F dy F dz m v dx v dy v dz dx dy dz ⎛⎞ =+ + = + + ⎜⎟ ⎝⎠ ∫∫ . 222222 2 2 tot 212121 2 1 11 1 (. 22 2 vv v x x y y z z Wm v d v v d v v d v m vvvvvv m v m v + = + + = ∫∫∫ EVALUATE: F " and d l " are vectors and have components. W and K are scalars and we never speak of their components.

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7-1 P OTENTIAL E NERGY AND E NERGY C ONSERVATION 7.1. IDENTIFY: grav Um g y = so grav 2 1 () g y y Δ= SET UP: y + is upward. EXECUTE: (a) 25 (75 kg)(9.80 m/s )(2400 m 1500 m) 6.6 10 J U = + × (b) (75 kg)(9.80 m/s )(1350 m 2400 m) 7.7 10 J U = − × EVALUATE: grav U increases when the altitude of the object increases. 7.2. IDENTIFY: Apply m = F a ! ! to the sack to find the force. cos WF s φ = . SET UP: The lifting force acts in the same direction as the sack&s motion, so 0 = ° EXECUTE: (a) For constant speed, the net force is zero, so the required force is the sack&s weight, 2 (5.00 kg)(9.80 m/s ) 49.0 N. = (b) (49.0 N) (15.0 m) 735 J W == . This work becomes potential energy. EVALUATE: The results are independent of the speed. 7.3. IDENTIFY: Use the free-body diagram for the bag and Newton’s first law to find the force the worker applies. Since the bag starts and ends at rest, 21 0 KK = and tot 0 W = . SET UP: A sketch showing the initial and final positions of the bag is given in Figure 7.3a. 2.0 m sin 3.5 m = and 34.85 = ° . The free-body diagram is given in Figure 7.3b. F !
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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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211_PartUniversity Physics Solution - 6-28 Chapter 6...

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