HW7-solutions

# HW7-solutions - kuehner (rjk679) – HW7 – Mackie –...

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Unformatted text preview: kuehner (rjk679) – HW7 – Mackie – (20211) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In an arcade game a 0 . 081 kg disk is shot across a frictionless horizontal surface by com- pressing it against a spring and releasing it. If the spring has a spring constant of 200 N / m and is compressed from its equi- librium position by 7 cm, find the speed with which the disk slides across the surface. Correct answer: 3 . 47833 m / s. Explanation: From conservation of mechanical energy, we have: 1 2 mv 2 f = 1 2 k x 2 i , v f = radicalBigg k x 2 i m = radicalBigg (200 N / m)(0 . 07 m) 2 . 081 kg = 3 . 47833 m / s . 002 10.0 points A 8 . 1 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 2 . 5 kg mass. 2 . 8 m ω 8 . 1 kg 2 . 5 kg Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 2 . 8 m . The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 5 . 38453 m / s. Explanation: Let : m 1 = 8 . 1 kg , m 2 = 2 . 5 kg , and ℓ = 2 . 8 m . Consider the free body diagrams 8 . 1 kg 2 . 5 kg T T m 1 g m 2 g a a Let the figure represent the initial config- uration of the pulley system (before m 1 falls down). From conservation of energy K i + U i = K f + U f 0 + m 1 g ℓ = m 2 g ℓ + 1 2 m 1 v 2 + 1 2 m 2 v 2 ( m 1- m 2 ) g ℓ = 1 2 ( m 1 + m 2 ) v 2 Therefore v = radicalBigg ( m 1- m 2 ) ( m 1 + m 2 ) 2 g ℓ = radicalBigg 8 . 1 kg- 2 . 5 kg 8 . 1 kg + 2 . 5 kg × radicalBig 2 (9 . 8 m / s 2 )(2 . 8 m) = 5 . 38453 m / s . keywords: 003 10.0 points Consider a bungee cord of unstretched length kuehner (rjk679) – HW7 – Mackie – (20211) 2 L = 40 m. When the cord is stretched to L > L it behaves like a spring and obeys Hooke’s law with the spring constant k = 40 N / m. However, unlike a spring, the cord folds instead of becoming compressed when the distance between its ends is less than the unstretched length: For L < L the cord has zero tension and zero elastic energy. To test the cord’s reliability, one end is tied to a high bridge (height H = 147 m above the surface of a river) and the other end is tied to a steel ball of weight mg = 100 kg × 9 . 8 m / s 2 . The ball is dropped off the bridge with zero initial speed. Fortunately, the cord works and the ball stops in the air before it hits the water — and then the cord pulls it back up. Calculate the ball’s height h bot at the lowest point of its trajectory. For simplicity, neglects the cord’s own weight and inertia as well as the air drag on the ball and the cord....
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## This note was uploaded on 03/24/2010 for the course PHYSICS 2202 taught by Professor Mihalisin during the Spring '09 term at Temple.

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HW7-solutions - kuehner (rjk679) – HW7 – Mackie –...

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