solution17

# solution17 - fullwood(tf5349 – Homework 8 Work and Energy...

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Unformatted text preview: fullwood (tf5349) – Homework 8 Work and Energy – Catala – (49033) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points An archer pulls her bowstring back 0 . 232 m by exerting a force that increases uniformly from zero to 342 N. What is the equivalent spring constant of the bow? Correct answer: 1474 . 14 N / m. Explanation: Given : F = 342 N and Δ x = 0 . 232 m . The equivalent spring constant is given by F = k Δ x k = F Δ x = 342 N . 232 m = 1474 . 14 N / m . 002 (part 2 of 2) 10.0 points How much work does the archer do in pulling the bow? Correct answer: 39 . 672 J. Explanation: From the work-kinetic energy theorem, W = 1 2 k (Δ x ) 2- = 1 2 (1474 . 14 N / m)(0 . 232 m) 2 = 39 . 672 J . 003 10.0 points A tugboat pulls a ship with a constant net horizontal force of 7 . 50 × 10 3 N and causes the ship to move through a harbor. How much work does the tugboat do on the ship if each moves a distance of 4.76 km ? Correct answer: 3 . 57 × 10 7 J. Explanation: Basic Concept: W net = F net d cos θ = F net d since θ = 0 ◦ ⇒ cos θ = 1. Given: F net = 7 . 50 × 10 3 N d = 4 . 76 km Solution: The force and the displacement are parallel, so the net work is W net = F net d = (7500 N)(4 . 76 km) · 1000 m 1 km = 3 . 57 × 10 7 J since 1 J = 1 N · m. 004 10.0 points A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 47 N. How far must the student be pushed, start- ing from rest, so that her final kinetic energy is 350 J ? Correct answer: 7 . 44681 m. Explanation: Basic Concepts: W net = Δ KE = KE f- KE i = KE f since v i = 0 m/s ⇒ KE i = 1 2 mv 2 i = 0 J. W net = F net d cos θ = F net d since θ = 0 ◦ ⇒ cos θ = 1. Given: F net = 47 N KE f = 350 J Solution: F net d = KE f d = KE f F net = 350 J 47 N = 7 . 44681 m fullwood (tf5349) – Homework 8 Work and Energy – Catala – (49033) 2 005 (part 1 of 2) 10.0 points A 1 . 39 × 10 3 kg car accelerates uniformly from rest to 12.9 m/s in 2.21 s. a) What is the work done on the car in this time interval?...
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solution17 - fullwood(tf5349 – Homework 8 Work and Energy...

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