Homework #6 Newton's 2nd Law-solutions

# Homework #6 Newton's 2nd Law-solutions - sanchez(es25396...

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Unformatted text preview: sanchez (es25396) – Homework #6 Newton’s 2nd Law – guevara – (130101) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A man weighs 439 N on Earth. What would he weigh on Jupiter, where the free-fall acceleration is 25 . 9 m / s 2 ? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1160 . 21 N. Explanation: The man’s weight on earth is given by W e = mg e m = W e g e = 439 N 9 . 8 m / s 2 = 44 . 7959 kg His weight on Jupiter will be W J = mg J = (44 . 7959 kg) (25 . 9 m / s 2 ) = 1160 . 21 N . 002 (part 1 of 3) 10.0 points A person stands on a scale in an elevator. The maximum and minimum scale readings are 637 . 2 N and 294 . 3 N, respectively. The acceleration of gravity is 9 . 8 m / s 2 . Assume the magnitude of the acceleration is the same during starting and stopping, and determine the weight of the person. Correct answer: 465 . 75 N. Explanation: We can express the weight read on the scale as the sum and difference of the person’s weight and the additional force of accelera- tion. F max = W + m a F min = W - m a Adding these two equations we obtain, F max + F min = 2 W Solving for W , W = F max + F min 2 = 637 . 2 N + 294 . 3 N 2 = 465 . 75 N . 003 (part 2 of 3) 10.0 points Determine the person’s mass. Correct answer: 47 . 5255 kg. Explanation: Since W = m g , the mass of the person is given by m = W g = 465 . 75 N 9 . 8 m / s 2 = 47 . 5255 kg . 004 (part 3 of 3) 10.0 points Determine the acceleration of the elevator....
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## This note was uploaded on 07/30/2009 for the course PHYS 1301 taught by Professor Guevara during the Spring '09 term at Texas A&M.

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Homework #6 Newton's 2nd Law-solutions - sanchez(es25396...

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