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2) As shown below, a passenger pulls on the suitcase that has a mass M, with a force of 30.0 N and displaces the suitcase horizontally 10.0 m. The angle between F and the horizontal is 30°. The coefficient of kinetic friction HK = 0.45 Force and displacement are not in the same direction; here W = FAx. Ar (b) The force that the passenger applies to the suitcase is not in the same direction as the displacement of the suitcase, so we have to find the "piece" or "component" of the force vector F that is in the same direction as the displacement. In other words, we must find F, = Fcos0, where 0 is the angle between the two vectors, F and Fx a. If 0 = 30°, what work is done on the suitcase by the passenger? b. What is the work done on the suitcase by friction? (remember that we can always express the force due to kinetic friction as follows: f = HgN; so now that we have the force we can find the work done by multiplying this force by the displacement. Be careful --- the normal force is NOT mg.

2) As shown below, a passenger pulls on the suitcase that has a mass M, with a force of 30.0 N and displaces the suitcase horizontally 10.0 m. The angle between F and the horizontal is 30°. The coefficient of kinetic friction HK = 0.45 Force and displacement are not in the same direction; here W = FAx. Ar (b) The force that the passenger applies to the suitcase is not in the same direction as the displacement of the suitcase, so we have to find the "piece" or "component" of the force vector F that is in the same direction as the displacement. In other words, we must find F, = Fcos0, where 0 is the angle between the two vectors, F and Fx a. If 0 = 30°, what work is done on the suitcase by the passenger? b. What is the work done on the suitcase by friction? (remember that we can always express the force due to kinetic friction as follows: f = HgN; so now that we have the force we can find the work done by multiplying this force by the displacement. Be careful --- the normal force is NOT mg.

Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter5: Newton's Laws Of Motion
Section: Chapter Questions
Problem 65PQ: A box with mass m1 = 6.00 kg sliding on a rough table with a coefficient of kinetic friction of...
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2) As shown below, a passenger pulls on the suitcase that has a mass M, with a force of 30.0 N and displaces the suitcase horizontally 10.0 m. The angle between F and the horizontal is 30°. The coefficient of kinetic friction HK = 0.45 Force and displacement are not in the same direction; here W = FAx. Ar (b) The force that the passenger applies to the suitcase is not in the same direction as the displacement of the suitcase, so we have to find the "piece" or "component" of the force vector F that is in the same direction as the displacement. In other words, we must find F, = Fcos0, where 0 is the angle between the two vectors, F and Fx a. If 0 = 30°, what work is done on the suitcase by the passenger? b. What is the work done on the suitcase by friction? (remember that we can always express the force due to kinetic friction as follows: f = HgN; so now that we have the force we can find the work done by multiplying this force by the displacement. Be careful --- the normal force is NOT mg.
Transcribed Image Text:2) As shown below, a passenger pulls on the suitcase that has a mass M, with a force of 30.0 N and displaces the suitcase horizontally 10.0 m. The angle between F and the horizontal is 30°. The coefficient of kinetic friction HK = 0.45 Force and displacement are not in the same direction; here W = FAx. Ar (b) The force that the passenger applies to the suitcase is not in the same direction as the displacement of the suitcase, so we have to find the "piece" or "component" of the force vector F that is in the same direction as the displacement. In other words, we must find F, = Fcos0, where 0 is the angle between the two vectors, F and Fx a. If 0 = 30°, what work is done on the suitcase by the passenger? b. What is the work done on the suitcase by friction? (remember that we can always express the force due to kinetic friction as follows: f = HgN; so now that we have the force we can find the work done by multiplying this force by the displacement. Be careful --- the normal force is NOT mg.
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