Question Part Points Submissions Used Suppose the ski patrol lowers a rescue sle
ID: 1904582 • Letter: Q
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Question Part Points Submissions Used Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 85.0 kg, down a ? = 66.0? slope at constant speed, as shown in Figure 6.22. The coefficient of friction between the sled and the snow is 0.100. Figure 6.22 (a) How much work is done by friction as the sled moves 30.0 m along the hill? J (b) How much work is done by the rope on the sled in this distance? J (c) What is the work done by gravity on the sled? J (d) What is the total work done? J
Question Part Points Submissions Used Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 85.0 kg, down a ? = 66.0? slope at constant speed, as shown in Figure 6.22. The coefficient of friction between the sled and the snow is 0.100. Figure 6.22 (a) How much work is done by friction as the sled moves 30.0 m along the hill? J (b) How much work is done by the rope on the sled in this distance? J (c) What is the work done by gravity on the sled? J (d) What is the total work done? JExplanation / Answer
A team of dogs drags a 43.7 kg sled 1.01 km over a horizontal surface at a constant speed. The coefficient of? A team of dogs drags a 43.7 kg sled 1.01 km over a horizontal surface at a constant speed. The coefficient of friction between the sled and the snow is 0.208. The acceleration of gravity is 9.8 m/s^2 . Find the work done by the dogs. Answer in units of kJ. Part B. Find the energy lost due to friction. Answer in units of kJ. Well we need to find the applied force: Fap = Fre + Fric resultant force, Fre, is mass x acceleration friction force, Fric, is mu x normal force normal force is the weight mu is the coefficient of friction Then use the applied force in The equation W=F*D And The energy lost is equal to Work (W=(delta) Ke so the answers are the same
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