231_PartUniversity Physics Solution

231_PartUniversity Physics Solution - 7-20 7.61. Chapter 7...

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7-20 Chapter 7 7.61. IDENTIFY and SET UP: There are two situations to compare: stepping off a platform and sliding down a pole. Apply the work-energy theorem to each. (a) EXECUTE: Speed at ground if steps off platform at height h : 1 1 other 2 2 K UW KU ++ = + 2 1 2 2 , mgh mv = so 2 2 2 vg h = Motion from top to bottom of pole: (take 0 y = at bottom) 1 1 other 2 2 K 2 1 2 2 mgd fd mv −= Use 2 2 2 h = and get mgd fd mgh () f dm g dh =− / ( 1 / ) f mg d h d mg h d = EVALUATE: For hd = this gives 0 f = as it should (friction has no effect). For 0, h = 2 0 v = (no motion). The equation for f gives f mg = in this special case. When f mg = the forces on him cancel and he doesn&t accelerate down the pole, which agrees with 2 0. v = (b) EXECUTE: 2 (1 / ) (75 kg)(9.80 m/s )(1 1.0 m/2.5 m) 441 N. fm g h d =−= = (c) Take 0 y = at bottom of pole, so 1 yd = and 2 . yy = 1 1 other 2 2 K 2 1 2 0( ) mgd f d y mv mgy +−− = + 2 1 2 mv mg d y f d y Using / ) f mg h d gives 2 1 2 ()( 1 / ) mv mg d y mg h d d y 2 1 2 (/) ( ) mv mg h d d y and 2( 1 /) h y d EVALUATE: This gives the correct results for 0 y = and for . = 7.62. IDENTIFY: Apply Eq.(7.14) to each stage of the motion. SET UP: Let 0 y = at the bottom of the slope. In part (a), other W is the work done by friction. In part (b), other W is the work done by friction and the air resistance force. In part (c), other W is the work done by the force exerted by the snowdrift. EXECUTE: (a) The skier&s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, 1 (60.0 kg)(9.8 N/kg)(65.0 m) 10,500 J, f Km g hW = or 1 38,200 J 10,500 J 27,720 J K . Then 1 1 2 2(27,720 J) 30.4 m/s 60 kg K v m == = . (b) 2 1 air k air ( ) 27,720 J ( ). f K KWW m g df d μ =− + = + 2 27,720 J [(0.2)(588 N)(82 m) (160 N)(82 m)] K + or 2 27,720 J 22,763 J 4957 J K . Then, 2 2 2(4957 J) 12.9 m/s 60 kg K v m = (c) Use the Work-Energy Theorem to find the force. , WK = Δ / (4957 J) (2.5 m) 2000 N FK d = . EVALUATE: In each case, other W is negative and removes mechanical energy from the system. 7.63. IDENTIFY and SET UP: First apply m = F a ! ! to the skier. Find the angle α where the normal force becomes zero, in terms of the speed v 2 at this point. Then apply the work-energy theorem to the motion of the skier to obtain another equation that relates v 2 and . Solve these two equations for . Let point 2 be where the skier loses contact with the snowball, as sketched in Figure 7.63a Loses contact implies 0. n 1 , yR = 2 cos = Figure 7.63a
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Potential Energy and Energy Conservation 7-21 First, analyze the forces on the skier when she is at point 2. The free-body diagram is given in Figure 7.63b. For this use coordinates that are in the tangential and radial directions. The skier moves in an arc of a circle, so her acceleration is 2 rad /, av R = directed in towards the center of the snowball. EXECUTE: y y Fm a = 2 2 cos / mg n mv R α −= But 0 n = so 2 2 cos / mg mv R = 2 2 cos vR g = Figure 7.63b Now use conservation of energy to get another equation relating 2 v to : 1 1 other 2 2 K UW KU ++ = + The only force that does work on the skier is gravity, so other 0. W = 1 0, K = 2 1 22 2 K mv = 11 , Um g ym g R == cos g g R Then 2 1 2 2 cos mgR mv mgR =+ 2 2 2( 1c o s ) vg R =− Combine this with the y y a = equation: cos 2 (1 cos ) Rg gR cos 2 2cos 3cos 2 = so cos 2/3 = and 48.2
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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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231_PartUniversity Physics Solution - 7-20 7.61. Chapter 7...

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