Wnet = m(v₁)² = m(v₁)², Wnet = Σ₁ W₁, W = Fd cos 0Problem 1:A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts witha speed of 5.5 m/sec, how far does is slide before coming to rest? Answer: d = 6.1 mProblem 2:What is the force a 60.0 kg sprinter exerts backward on the track to accelerate from 2.00 m/secto 8.00 m/sec in a distance of 25.0 m, if they encounter a wind that exerts an average force of30.0 N against them?Hint: Wnet = Wrunner - Wwind. Answer: Frunner: = 102.0 NProblem 3:A 500-kg dragster accelerates from rest to a final speed of 110 m/sec in 400 m (about a quarterof a mile) and encounters an average frictional force of 1200 N. What is the work done by thedragster? Hint: Wnet = Wdragster - Wfriction. Answer: Wdragster = 3.5 × 106 J

Name: Wnet = m(v₁)² = m(v₁)², Wnet = Σ₁ W₁, W = Fd cos 0Problem 1:A 2.0 kg
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Description: Wnet = m(v₁)² = m(v₁)², Wnet = Σ₁ W₁, W = Fd cos 0Problem 1:A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts witha speed of 5.5 m/sec, how far does is slide before coming to rest? Answer: d = 6.1 mProbl

Mass of box, Friction force on the box, The initial speed of…

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Problem 1: A 2.0 kg box slides on a floor. A friction force of 5.0 N opposes the motion. If the box starts with a speed of 5.5 m/sec, how far does is slide before coming to rest? Answer: d = 6.1 m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter5: Energy

Section: Chapter Questions

Problem 49P: An 80.0-kg skydiver jumps out of a balloon at an altitude of 1.00 103 m and opens the parachute at...

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