Question

GOAL Apply the basic definitions of work done by a constant 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 of 1.20 102 N on the sled by pulling on the rope. (a) How much work does he do on the sled if the rope is horizontal to the ground (θ = 0° in the figure) and he pulls the sled 5.00 m? (b) How much work does he do on the sled if θ = 30.0° and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.) (c) At a coordinate position of 12.4 m, the Eskimo lets up on the applied force. A friction force of 45.0 N between the ice and the sled brings the sled to rest at a coordinate position of 18.2 m. How much work does friction do on the sled? STRATEGY Substitute the given values of F and Δx into the basic equations for work. SOLUTION (A) Find the work done when the force is horizontal. Use the proper equation, substituting the given values. W = FxΔx = (1.20 102 N)(5.00 m) = 6.00 102 J (B) Find the work done when the force is exerted at a 30° angle. Use the proper equation, again substituting the given values. W = (F cos θ) d = (1.20 102 N)(cos 30.0°)(5.00 m) = 5.20 102 J (C) How much work does a friction force of 45.0 N do on the sled as it travels from a coordinate position of 12.4 m to 18.2 m? Use the proper equation, with Fx replaced by fk. Wfric = FxΔx = fk(xf − xi) Substitute fk = −45.0 N and the initial and final coordinate positions into xi and xf. Wfric = (−45.0 N)(18.2 m − 12.4 m) = −2.6 102 J LEARN MORE REMARKS The normal force n with arrow, the gravitational force mg with arrow, and the upward component of the applied force do no work on the sled because they're perpendicular to the displacement. The mass of the sled didn't come into play here, but it is important when the effects of friction must be calculated and in the work-energy theorem. QUESTION How does the answer for the work done by the applied force change if the load is doubled? The work is doubled. The work is halved. The work is the same. Explain. This answer has not been graded yet. PRACTICE IT Use the worked example above to help you solve this 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 of 1.00 102 N on the sled by pulling on the rope. (a) How much work does he do on the sled if the rope is horizontal to the ground (θ = 0° in the figure) and he pulls the sled 5.80 m? J (b) How much work does he do on the sled if θ = 30.0° and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.) J (c) At a coordinate position of 5.80 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.80 m. How much work does friction do on the sled? J EXERCISE HINTS: GETTING STARTED | I'M STUCK! Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 53.0 N. (a) If he does 3.10 102 J of work on the sled while exerting the force horizontally, through what distance must he have pushed it? m (b) If he exerts the same force at an angle of 35.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled? J

Answer 1