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Unformatted text preview: pampalone (tp8327) – HW#11 – swaminathan – (10500) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 3 . 41 kg ball is attached to a 13 . 3 N fish line. The ball is released from rest at the horizontal position ( θ = 90 ◦ ). At what angle θ (measured from the verti cal) will the fish line break? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 82 . 3765 ◦ . Explanation: In the swing down to the breaking point, energy is conserved: mg r cos θ = mv 2 2 2 g m cos θ = mv 2 r . At the breaking point, consider the radial forces: summationdisplay F r = ma r T mg cos θ = mv 2 r T = 3 mg cos θ θ = arccos parenleftbigg T 3 mg parenrightbigg = arccos bracketleftbigg 13 . 3 N 3(3 . 41 kg)(9 . 8 m / s 2 ) bracketrightbigg = 82 . 3765 ◦ . 002 (part 1 of 2) 10.0 points A ball of mass m is attached to a spring of negligible mass, and force constant k . The spring is freely pivoted at the opposite end from the ball, and the system is initially at rest and horizontal. L L +Δ L If the ball is allowed to fall, what is its speed when the system is swinging through the vertical and the spring is stretched an amount Δ L = L 10 , where L is the spring’s initial, unstretched length? 1. radicalBigg parenleftbigg k L m 11 g parenrightbiggparenleftbigg L 10 parenrightbigg 2. radicalBigg parenleftbigg k L 10 m 22 g parenrightbigg L 10 3. radicalBigg parenleftbigg 10 g k L m parenrightbiggparenleftbigg L 100 parenrightbigg 4. radicalBigg k L 2 mg 5. Not enough information is given. 6. radicalbigg mg L 10 7. radicalBigg parenleftbigg k L m 10 g parenrightbigg L 8. radicalBigg parenleftbigg g k L m parenrightbigg L 9. radicalBigg parenleftbigg 22 g k L 10 m parenrightbigg L 10 correct Explanation: Using conservation of energy with E = K + U s + U g and taking U g = 0 at the instant the system is vertical we get E i = E f pampalone (tp8327) – HW#11 – swaminathan – (10500) 2 K i + U si + U gi = K f + U sf + U gf 0 + 0 + mg ( L + Δ L ) = 1 2 mv 2 + 1 2 k (Δ L ) 2 mg parenleftbigg 11 L 10 parenrightbigg = 1 2 mv 2 + 1 2 k parenleftbigg L 2 100 parenrightbigg 22 mg L 10 k L 2 100 = mv 2 parenleftbigg mL 10 parenrightbiggparenleftbigg 22 g k L 10 m parenrightbigg = mv 2 radicalBigg parenleftbigg 22 g k L 10 m parenrightbigg L 10 = v 003 (part 2 of 2) 10.0 points Having calculated the speed in the first part, now find the acceleration a of the ball at the point where it swings through the vertical position. 1. ˆ j parenleftbigg k L 10 m g parenrightbigg correct 2. ˆ j parenleftBigg g v 2 11 L 10 parenrightBigg 3. ˆ j parenleftbigg k L 10 m g parenrightbigg 4....
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 Spring '08
 KEN
 Energy, Force, Friction, Mass, Potential Energy, Wnet

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