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A box of mass 15 kg was pushed by a person who has let go of it so that it is currently moving at 55 m/s across a rough surface. Find the amount of friction exerted on the box if it stops after moving 3.15 m. v = 55 m/s v = 0 m/s 15 kg 3.15 m

A box of mass 15 kg was pushed by a person who has let go of it so that it is currently moving at 55 m/s across a rough surface. Find the amount of friction exerted on the box if it stops after moving 3.15 m. v = 55 m/s v = 0 m/s 15 kg 3.15 m

Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter5: The Laws Of Motion
Section: Chapter Questions
Problem 5.41P: Figure P5.41 shows the speed of a persons body as he does a chin-up. Assume the motion is vertical...
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Problems 1 and 2 were done incorrectly. Correct the mistakes made and find the real solutions to the problems. Find the solution to Problem 3.

1. A box of mass 15 kg was pushed by a person who has let go of it so that it is currently moving at 55 m/s across a rough surface. Find the amount of friction exerted on the box if it stops after moving 3.15 m. v = 55 m/s v = 0 m/s 15 kg 3.15 m W. "person E, E, E, person box person box floor floor Earth Earth's field iion E + W, person friction ½ mv? + Epesoa Ax = F, Ax Ensson = ma = (15)(55)% 3D 825 N 2 (15)(55)? + 825(3.15) = F, (3.15) 25,286.25 = F, (3.15) F; = 8,027 N
Transcribed Image Text:1. A box of mass 15 kg was pushed by a person who has let go of it so that it is currently moving at 55 m/s across a rough surface. Find the amount of friction exerted on the box if it stops after moving 3.15 m. v = 55 m/s v = 0 m/s 15 kg 3.15 m W. "person E, E, E, person box person box floor floor Earth Earth's field iion E + W, person friction ½ mv? + Epesoa Ax = F, Ax Ensson = ma = (15)(55)% 3D 825 N 2 (15)(55)? + 825(3.15) = F, (3.15) 25,286.25 = F, (3.15) F; = 8,027 N
2. A skydiver dropped a block of wood at the same time she dove out of the plane. The skydiver has a mass of 85 kg and the block 15 kg. After falling 10 m, the block and the skydiver have the same speed (we are neglecting air resistance while the parachute has not been used). The skydiver then opens her parachute and, 10 additional meters later, the skydiver is moving at 16 m/s. a. How much faster is the block than the skydiver in this final state? b. How much energy was lost by the skydiver due to friction during these last 10 meters? c. Calculate the average air resistance (force) exerted on the skydiver. makydiver = 85 kg miock" 15 kg A Vaydiver = Valock = 0 air 10 m skydiver block B Earth's field skydiver block 10 m 16 m/s = v skydiver Vtiock Part a: E, E, E, Ega = Ega + E, mgh, = mgh, + ½ mv gh, = gh, + % ? (10)(20) = (10)(10) + ½ ? 200 = 100 + ½ 100 = ½ 200 = v2 v = 14.1 m/s g.8 block air Earth A в So the block is moving 1.9 m/s faster than the skydiver!
Transcribed Image Text:2. A skydiver dropped a block of wood at the same time she dove out of the plane. The skydiver has a mass of 85 kg and the block 15 kg. After falling 10 m, the block and the skydiver have the same speed (we are neglecting air resistance while the parachute has not been used). The skydiver then opens her parachute and, 10 additional meters later, the skydiver is moving at 16 m/s. a. How much faster is the block than the skydiver in this final state? b. How much energy was lost by the skydiver due to friction during these last 10 meters? c. Calculate the average air resistance (force) exerted on the skydiver. makydiver = 85 kg miock" 15 kg A Vaydiver = Valock = 0 air 10 m skydiver block B Earth's field skydiver block 10 m 16 m/s = v skydiver Vtiock Part a: E, E, E, Ega = Ega + E, mgh, = mgh, + ½ mv gh, = gh, + % ? (10)(20) = (10)(10) + ½ ? 200 = 100 + ½ 100 = ½ 200 = v2 v = 14.1 m/s g.8 block air Earth A в So the block is moving 1.9 m/s faster than the skydiver!
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