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Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals, for it can go from rest to 33.5 m/s in 4.41 s. If its mass is 117 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower. (a) Number (b) Number i Units Units V

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals, for it can go from rest to 33.5 m/s in 4.41 s. If its mass is 117 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower. (a) Number (b) Number i Units Units V

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Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals, for it can go from rest to 33.5 m/s in 4.41 s. If its mass is 117 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower. (a) Number i (b) Number i Units Units
Transcribed Image Text:Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals, for it can go from rest to 33.5 m/s in 4.41 s. If its mass is 117 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower. (a) Number i (b) Number i Units Units
ANIMATED FIGURE 6.20 (a) A compound bow. (b) A plot of F cos versus s as the bowstring is drawn back. When the force varies with the displacement, as in Animated Figure 6.20b, we cannot use the relation W= (F cos 0)s to find the work, because this equation is valid only when the force is constant. However, we can use a graphical method. In this method we divide the total displacement into very small segments, As₁, As, and so on (see Figure 6.21a). For each segment, the average value of the force component is indicated by a short horizontal line. For example, the short horizontal line for segment As, is labeled (F cos ), in Figure 6.21a. We can then use this average value as the constant-force component in Equation 6.1 and determine an approximate value for the work AW₁ done during the first segment: AW₁ = (F cos 0)₁AS₁. But this work is just the area of the colored rectangle in the drawing. The word "area" here refers to the area of a rectangle that has a width of As, and a height of (F cos 0)₁; it does not mean an area in square meters, such as the area of a parcel of land. In a like manner, we can calculate an approximate value for the work for each segment. Then we add the results for the segments to get, approximately, the work W done by the variable force: F cos 0 W (F cos 0)₁ As₁ + (F cos 0)₂ As2 + ... (F cos 0)₂ (F cos 0)₁ Δ1 Δ2 483 484 (a) F cos e Work
Transcribed Image Text:ANIMATED FIGURE 6.20 (a) A compound bow. (b) A plot of F cos versus s as the bowstring is drawn back. When the force varies with the displacement, as in Animated Figure 6.20b, we cannot use the relation W= (F cos 0)s to find the work, because this equation is valid only when the force is constant. However, we can use a graphical method. In this method we divide the total displacement into very small segments, As₁, As, and so on (see Figure 6.21a). For each segment, the average value of the force component is indicated by a short horizontal line. For example, the short horizontal line for segment As, is labeled (F cos ), in Figure 6.21a. We can then use this average value as the constant-force component in Equation 6.1 and determine an approximate value for the work AW₁ done during the first segment: AW₁ = (F cos 0)₁AS₁. But this work is just the area of the colored rectangle in the drawing. The word "area" here refers to the area of a rectangle that has a width of As, and a height of (F cos 0)₁; it does not mean an area in square meters, such as the area of a parcel of land. In a like manner, we can calculate an approximate value for the work for each segment. Then we add the results for the segments to get, approximately, the work W done by the variable force: F cos 0 W (F cos 0)₁ As₁ + (F cos 0)₂ As2 + ... (F cos 0)₂ (F cos 0)₁ Δ1 Δ2 483 484 (a) F cos e Work
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