EXAMPLE 5.2 More Sledding GOAL Calculate the work done by friction when an objec
ID: 1774082 • Letter: E
Question
EXAMPLE 5.2 More Sledding GOAL Calculate the work done by friction when an object is acted on by an applied force. PROBLEM An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force on the sled by pulling on the rope. Suppose the coefficient of kinetic friction between the loaded sled and snow is 0.200. (a) The Eskimo pulls the sled 5.00 m, exerting a force of 1.20 x 10 N at an angle of 0°. Find the work done on the sled by friction, Antontal pu ng . and the net work. (Note the work done by the applied force alone would be 6.00 x 102 in this situation.) (b) Repeat the calculation if the applied force is exerted at an angle of 30.00 with the horizontal. (Note the work done by the applied force alone would be 5.20 x 102 J in this situation.) STRATEGY See the figure. The frictional work depends on the magnitude of the kinetic friction coeficient, the normal force, and the displacement. Use the y component of Newton's second law to find the normal force ni, calculate the work done by friction using the definitions, and sum with the frictionless result given above to obtain the net work on the sled. Part (b) is solved similarly, but the normal force is smaller because it has the help of the applied force Fapp in supporting the load. SOLUTION (A) Find the work done by friction on the sled and the net work, if the applied force is horizontal. First, find the norm ,-n-mg-o-n-mg nal force from they component of Newton's second law, which involves only the normal force and the force of gravity Use the normal force to compute the work done by friction. (0.200)(50.0 kg)(9.80 m/s2)(5.00 m) -4.90 x 10 Sum the frictional work with the work done by the applied force to get the net work 6.00 x 10(4.90 x 10)o+0 1.10 x 10 jExplanation / Answer
Practice it
a)
Wnet = 110x 5.9 -578.2
Wnet = 70.8 J
b)
Wfriction = -ux 5.9 ( mg - Fsin 30 )
Wf = -513.3 J
Wnet = 5.9xcos 30 110 - 513.3
Wnet = 48.75 J
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