Homework #8 - Conservation of energy and linear momentum-solutions

# Homework #8 - Conservation of energy and linear momentum-solutions

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Unformatted text preview: sanchez (es25396) Homework #8 - Conservation of energy and linear momentum guevara (130101) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Does the KE of a car change more when it accelerates from 22 km / h to 32 km / h or when it accelerates from 32 km / h to 42 km / h? 1. No difference 2. From 32 km / h to 42 km / h correct 3. From 22 km / h to 32 km / h 4. More information is needed. Explanation: The change in kinetic energy is 1 2 mv 2 f- 1 2 mv 2 i , so we are comparing v 2 f- v 2 i . (42 km / h) 2- (32 km / h) 2 = 740 km 2 / h 2 is larger than (32 km / h) 2- (22 km / h) 2 = 540 km 2 / h 2 , so KE changes more for a given v at the higher speeds. 002 (part 1 of 4) 10.0 points A 6 kg block slides down a frictionless incline making an angle of 48 with the horizontal. The acceleration of gravity is 9 . 81 m / s 2 . a) Find the work done by the gravitational force when the block slides 1 . 5 m (measured along the incline). Correct answer: 65 . 6123 J. Explanation: Let : m = 6 kg , = 48 , d = 1 . 5 m , and g = 9 . 81 m / s 2 . N mg block The normal force does not contribute to the work since it is perpendicular to the displace- ment of the block, so W = F x d = mg sin d = (6 kg) (9 . 81 m / s 2 ) sin48 (1 . 5 m) = 65 . 6123 J . 003 (part 2 of 4) 10.0 points b) What is the total work done on the block? Correct answer: 65 . 6123 J. Explanation: Here the total force that acts on the block is only F x since the component F y cancels out with the normal force so W tot = F x d = (6 kg) (9 . 81 m / s 2 ) sin 48 (1 . 5 m) = 65 . 6123 J . 004 (part 3 of 4) 10.0 points c) What is the speed of the block after it has moved 1 . 5 m if it starts from rest? Correct answer: 4 . 67662 m / s. Explanation: By conservation of energy, K = W tot 1 2 m ( v 2 f- 0) = W tot v f = radicalbigg 2 W tot m = radicalBigg 2 (65 . 6123 J) 6 kg = 4 . 67662 m / s . sanchez (es25396) Homework #8 - Conservation of energy and linear momentum guevara (130101) 2 005 (part 4 of 4) 10.0 points d) What is its speed after 1 . 5 m if it starts with an initial speed of 1 . 5 m / s? Correct answer: 4 . 91129 m / s....
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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 #8 - Conservation of energy and linear momentum-solutions

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