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Page 138 Practice Problem 5.14: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm.

Page 138 Practice Problem 5.14: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm.

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.
Chapter14: Static Equilibrium, Elasticity, And Fracture
Section: Chapter Questions
Problem 70PQ
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140 CHAPTER 5 Applications of Newton's Laws EXAMPLE 5.14 Fishy business In this example we will consider the forces that come into play in a spring balance. Suppose that a spring balance used to weigh fish is built with a spring that stretches 1.00 cm when a 12.0 N weight is placed in the pan. When the 12.0 N weight is replaced with a 1.50 kg fish, what distance does the spring stretch? SOLUTION SET UP Figure 5.22a and b shows how the spring responds to the 12.0 N weight and to the 1.50 kg fish. The spring and fish are in equi- librium. We draw a free-body diagram for the fish (Figure 5.22c), show- ing the downward force of gravity and the upward force exerted by the stretched spring. We point the y axis vertically upward, as usual. SOLVE We want to use Hooke's law (Fspr= kAL) to relate the stretch of the spring to the force, but first we need to find the force constant k. The problem states that forces of magnitude 12.0 N are required to stretch the spring AL = 1.00 cm (= 1.00 × 10-2 m), so bo 1200 N/m. k= 12.0 N 1.00 × 102 m ► FIGURE 5.22 Weighing a fish. the che AL= 1.00 cm adi Charson22 svitie --X 12.0 N (a) The scale stretched by a known weight The weight of the 1.50 kg fish is w = mg = (1.50 kg) (9.80 m/s²) The equilibrium condition for the fish is 50169 31 115) and finally, AL = 0.0123 m = 1.23 cm. REFLECT The weight of the fish is greater than the amount of force needed to stretch the spring 1.00 cm, so we expect the fish to stretch the spring more than 1.00 cm. -mg + Fspr = 0, ΣΕ, = 0, -14.7 N+ (1200 N/m) AL = 0, Practice Problem: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm. ute etter AL=" 38380 ///////////z m = 1.50 kg w = mg Video Tutor Solut = 14.7 N. (b) The scale stretched by a known mass Fspr W = mg (c) Free-body diagram for the fish REF are i is th that 5.5 Th tu fr u
Transcribed Image Text:140 CHAPTER 5 Applications of Newton's Laws EXAMPLE 5.14 Fishy business In this example we will consider the forces that come into play in a spring balance. Suppose that a spring balance used to weigh fish is built with a spring that stretches 1.00 cm when a 12.0 N weight is placed in the pan. When the 12.0 N weight is replaced with a 1.50 kg fish, what distance does the spring stretch? SOLUTION SET UP Figure 5.22a and b shows how the spring responds to the 12.0 N weight and to the 1.50 kg fish. The spring and fish are in equi- librium. We draw a free-body diagram for the fish (Figure 5.22c), show- ing the downward force of gravity and the upward force exerted by the stretched spring. We point the y axis vertically upward, as usual. SOLVE We want to use Hooke's law (Fspr= kAL) to relate the stretch of the spring to the force, but first we need to find the force constant k. The problem states that forces of magnitude 12.0 N are required to stretch the spring AL = 1.00 cm (= 1.00 × 10-2 m), so bo 1200 N/m. k= 12.0 N 1.00 × 102 m ► FIGURE 5.22 Weighing a fish. the che AL= 1.00 cm adi Charson22 svitie --X 12.0 N (a) The scale stretched by a known weight The weight of the 1.50 kg fish is w = mg = (1.50 kg) (9.80 m/s²) The equilibrium condition for the fish is 50169 31 115) and finally, AL = 0.0123 m = 1.23 cm. REFLECT The weight of the fish is greater than the amount of force needed to stretch the spring 1.00 cm, so we expect the fish to stretch the spring more than 1.00 cm. -mg + Fspr = 0, ΣΕ, = 0, -14.7 N+ (1200 N/m) AL = 0, Practice Problem: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm. ute etter AL=" 38380 ///////////z m = 1.50 kg w = mg Video Tutor Solut = 14.7 N. (b) The scale stretched by a known mass Fspr W = mg (c) Free-body diagram for the fish REF are i is th that 5.5 Th tu fr u
Page 138 Practice Problem 5.14: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm.
Transcribed Image Text:Page 138 Practice Problem 5.14: If, instead of the fish, we place a 3.00 kg rock in the pan, how much does the spring stretch? Answer: twice as much, 2.45 cm.
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