Physics Solution Manual for 1100 and 2101

We will use this principle to determine the distance

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Unformatted text preview: – (71.1 N)(2.13 m – 1.52 m) = –43 J b. The change in potential energy is ∆PE = mghf – mgh0 = (71.1 N)(2.13 m – 1.52 m) = +43 J ______________________________________________________________________________ 33. REASONING The work done by the weight of the basketball is given by Equation 6.1 as W = ( F cos θ ) s , where F = mg is the magnitude of the weight, θ is the angle between the weight and the displacement, and s is the magnitude of the displacement. The drawing shows that the weight and displacement are parallel, so that θ = 0°. The potential energy of the mg basketball is given by Equation 6.5 as PE = mgh, where h is the height of the ball above the ground. SOLUTION a. The work done by the weight of the basketball is W = ( F cos θ ) s = mg (cos 0°)(h0 − hf) = (0.60 kg)(9.80 m/s )(6.1 m − 1.5 m) = 27 J 2 b. The potential energy of the ball, relative to the ground, when it is released is s Chapter 6 Problems 2 PE0 = mgh0 = (0.60 kg)(9.80 m/s )(6.1 m) = 36 J 303 (6.5) c. The potential energy of the ball, relative to the ground, when it is caught is PEf = mghf = (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8 J (6.5) d. The change in the ball’s gravitational potential energy is ∆PE = PEf − PE0 = 8.8 J – 36 J = −27 J We see that the change in the gravitational potential energy is equal to –27 J = −W , where W is the work done by the weight of the ball (see part a). ______________________________________________________________________________ 34. REASONING The gravitational force on Rocket Man is conservative, but the force generated by the propulsion unit is nonconservative. Therefore, the work WP done by the propulsive force is equal to the net work done by all external nonconservative forces: WP = Wnc. We will use Wnc = Ef − E0 (Equation 6.8) to calculate Wnc in terms of Rocket Man’s initial mechanical energy and final mechanical energy. Because Rocket Man’s speed and height increase after he leaves the ground, both his kinetic energy KE = 1 mv 2 2 (Equation 6.2) and potential energy PE = mgh (Equation 6.5) increase. Therefore, his final mechanical energy Ef = KEf + PEf is greater than his initial mechanical energy E0 = KE0 + PE0, and the work WP = Wnc = Ef − E0 done by the propulsive force is positive. SOLUTION We make the simplifying assumption that Rocket Man’s initial height is h0 = 0 m. Thus, his initial potential energy PE 0 = mgh0 is zero, and his initial kinetic energy 2 KE 0 = 1 mv0 is also zero (because he starts from rest). This means he has no initial 2 mechanical energy, and we see that the work done by the propulsive force is equal to his final mechanical energy: WP = Ef − 0 = Ef . Expressing Rocket Man’s final mechanical energy Ef in terms of his final kinetic energy and his final potential energy, we obtain the work done by the propulsive force: WP = 1 mvf2 + mghf 2 ( ) = 1 (136 kg )( 5.0 m/s ) + (136 kg ) 9.80 m/s2 (16 m ) = +2.3 ×104 J 2 2 304 WORK AND ENERGY 35. SSM REASONING AND SOLUTION a. The work done by non-conservative forces is given by Equation 6.7b as Wnc = ∆KE + ∆PE ∆PE = Wnc – ∆KE so Now ∆KE = 1 2 mvf2 − 1 2 2 mv0 = 1 2 (55.0 kg)[(6.00 m/s) 2 − (1.80 m/s)2 ] = 901 J and ∆PE = 80.0 J – 265 J – 901 J = −1086 J b. ∆PE = mg (h – h0) so h – h0 = ∆ PE –1086 J = = –2.01 m mg (55.0 kg ) 9.80 m/s 2 ( ) This answer is negative, so that the skater’s vertical position has changed by 2.01 m and the skater is below the starting point . ______________________________________________________________________________ 36. REASONING The girl’s gravitational potential energy PE = mgh (Equation 6.5) increases as she rises. This increase comes at the expense of her kinetic energy KE = 1 mv 2 2 (Equation 6.2), which decreases as she rises. Because air resistance is negligible, all of the kinetic energy she loses is transformed into potential energy, so that her total mechanical energy E = K E + PE remains constant. We will use this principle to determine the distance she rises during this interval. SOLUTION The conservation principle gives KE f + PE f = KE 0 + PE 0 14 3 14 24 24 3 Ef or PE f − PE 0 = KE 0 − KE f or mg ( hf − h0 ) = KE 0 − KE f E0 Solving this expression for hf − h0 , which is the distance she rises, we obtain hf − h0 = 37. KE0 − KEf mg = 440 J − 210 J ( 35 kg ) ( 9.80 m/s2 ) = 0.67 m SSM REASONING No forces other than gravity act on the rock since air resistance is being ignored. Thus, the net work done by nonconservative forces is zero, Wnc = 0 J. Chapter 6 Problems 305 Consequently, the principle of conservation of mechanical energy holds, so the total mechanical energy remains constant as the rock falls. If we take h = 0 m at ground level, the gravitational potential energy at any height h is, according to Equation 6.5, PE = mgh . The kinetic energy of the rock is given by 2 1 Equation 6.2: KE = mv 2 . In order to use Equation 6.2, we must have a value for v at 2 2 each desired height h. The quantity v...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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