Question ( ) Solution. Approximate length of an atomic bond for solid aluminum.

Transcription

1 Question ( ) Approximate length of an atomic bond for solid aluminum. Spring A of stiness 10 N/m is hung vertically from a ringstand. From the lower end of this spring, a dierent spring (Spring B) of stiness 20 N/m is hung. And from the lower end of this spring, a 0.25-kg block is hung in equilibrium. When springs are connected in this orientation, they are said to be in series. Neglect the mass of the springs. (a) How far is Spring A stretched from its relaxed length? (b) How far is Spring B stretched from its relaxed length? (c) If you were to replace the set of springs with a single spring that would stretch the same amount when the block is hung from it, what is the stiness of the single spring? Solution (a) Begin by drawing a picture with the springs relaxed and with the springs stretched. Because Spring A is less sti than Spring B, it will stretch farther than Spring B. Figure 1: Two dierent springs connected in series. Dene Spring B and the block as the system. Draw a free-body diagram for the system. Apply the momentum principle. Because the system is in equilibrium, the change in momentum of the system is zero. F net = p t F net = 0

2 Figure 2: The system is dened as the block and Spring B. Figure 3: Free-body diagram for the system consisting of the block and Spring B. The forces on the system are the downward gravitational force and the upward force by Spring A. Because Spring A's mass is negligible, the mass of the system is composed of the mass of the block. Applying the denition of net force gives F net = F by spring A on system + F grav by Earth on system 0 = F by spring A on system + F grav by Earth on system F by spring A on system = F grav by Earth on system < 0, k A s A, 0 > = < 0, m block g, 0 > < 0, k A s A, 0 > = < 0, m block g, 0 > Write the above equation in component form for the y-component and solve for the distance stretched of Spring A.

3 k A s A = m block g s A = m blockg k A s A = 10 N/m s A = m Spring A stretches m. (b) To nd the distance the Spring B stretches, dene the block as the system and apply the momentum principle to the block. A free-body diagram is shown below. Figure 4: Free-body diagram for the block. The net force on the block is zero since the block is in equilibrium. 0 = F by spring B on block + F grav by Earth on block F by spring B on block = F grav by Earth on block < 0, k B s B, 0 > = < 0, m block g, 0 > < 0, k B s B, 0 > = < 0, m block g, 0 > Write the above equation in component form for the y-component. k B s B = m block g s B = m blockg k B s B = 20 N/m s B = m Spring B stretches m which is less than the distance stretched by Spring B, as expected. In fact, it's half as much. Since Spring B has twice the stiness, it stretches half the amount, because the force applied

4 to the end of each spring is the same. The total distance stretched by the two springs is m, which is 0.37 m rounded to two signicant gures. (c) If you replace the two springs with a single spring that will stretch the same amount when holding the block, then the single spring would stretch m, the same amount as the total distance stretched by the two spring together, as shown below. Figure 5: A single spring that stretches the same amount as the other two springs together. The single spring stretches more than either spring stretched alone. Thus, this single spring would be less sti than either spring. This is an important observation. Two springs in series are less sti than either spring alone. Applying the momentum principle to the block hanging from the single spring gives k eff s = m block g k eff = m blockg s k eff = m k eff = 6.67 N m To two signicant gures, it stiness is 6.7 m. This is less sti than either spring, exactly as expected, which is why it stretches more than either spring stretches. Another way to calculate the eective spring stiness of two springs in series is: 1 k eff = 1 k k 2 Substituting the stiness of the two springs and solving for the eective stiness gives

5 k eff = ( 1 10 N/m N/m = 6.67 N/m ) 1 which is in agreement with what we calculated earlier by applying the momentum principle and determining the total distance stretched.