3 2 40 pts use for parts b and c tarzan mass 70 kg

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2. (40 pts) Use for parts (b) and (c): Tarzan (mass 70 kg) stands on a ledge holding a 30-kg sack of fruit in one hand and the end of a 4-m long vine in the other. He swings down the vine to the right. At the bottom of the swing, when the vine is momentarily vertical and Tarzan is at ground level, he throws the sack to the right along the frictionless ground. Immediately after releasing the sack, Tarzan continues swinging to the right at 3 m/s. Use for part (a): The 30-kg sack, meanwhile, begins to slide up a gravelly hill ( μ k = 0 . 3) inclined at 20 deg to the horizontal, upward and to the right. As the sack slides up the hill, it compresses a spring with force constant 150 N/m. The sack comes to a stop after sliding a distance of 2.5 m up the hill, and therefore compressing the spring 2.5 m from its unstretched length. (a) What is the speed of the sack at the bottom of the hill? (b) What is the height of the ledge above the ground? Use the first paragraph, along with the answer from part (a). (c) As Tarzan and the sack swing downward, consider the instant the vine makes an angle of 25 deg with the vertical. What is the tension in the vine at this instant? (a) Use the work-energy equation to find the speed of the sack at the bottom of the hill: W friction = - μ k Nd = - μ k ( mg cos ) d U f + K f = W friction + U i + K i m s gd sin + 1 2 kd 2 + 0 = - μ k ( m s g cos ) d + 0 + 1 2 m s v 2 s 4
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v s = s 2 m s m s gd (sin + μ k cos ) + 1 2 kd 2 = s 2 30 (30)(9 . 8)(2 . 5)(sin(20) + (0 . 3) cos(20)) + 1 2 (150)(2 . 5) 2 = 7 . 86 m / s (b) We first use conservation of momentum to find the speed of Tarzan and the sack just before he throws it: p i = p f ( m T + m s ) v sT = m s v s + m T v T v sT = m s v s + m T v T m T + m s = (30)(7 . 86) + (70)(3) 70 + 30 = 4 . 46 m / s We now use energy conservation to determine the height of the ledge, where y = 0 is defined to be at ground level: U i + K i = U f + K f ( m s + m T ) gh + 0 = 1 2 ( m s + m T ) v 2 sT h = v 2 sT 2 g = 4 . 46 2 (2)(9 . 8) = 1 . 01 m (c) When the vine makes a 25-deg angle with the vertical, Tarzan’s height is given by h 2 = L (1 - cos 2 ) = (4)[1 - cos(25)] = 0 . 375 m We use energy conservation to determine Tarzan’s speed at this height: U 2 + K 2 = U i + K i ( m s + m T ) gh 2 + 1 2 ( m s + m T ) v 2 2 = ( m s + m T ) gh + 0 v 2 = p 2 g ( h - h 2 ) = p (2)(9 . 8)(1 . 01 - 0 . 375) = 3 . 54 m
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