道彥取一個彈性限度標示為100 gw,原長為10 cm的彈簧,測量20 gw的砝碼時,彈簧全長為14 cm;若將砝碼取下後,彈簧恢復為10 cm。 現在道彥在此彈簧下改掛一個玩具小熊(仍不超過彈性限度),則彈簧全長為18 cm。試問玩具小熊的重量為多少gw?
解答:This is a Hooke's Law problem. Hooke's Law states that the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance—that is, F s =kx, where k is a constant factor characteristic of the spring (its stiffness), and x is the displacement from the spring's equilibrium position. Here's how to solve the problem: 1. Determine the spring constant (k): Original length (natural length) of the spring: L 0 =10 cm Force applied (weight of the first mass): F 1 =20 gw Full length with 20 gw: L 1 =14 cm Extension (x 1 ) with 20 gw: x 1 =L 1 −L 0 =14 cm−10 cm=4 cm Using Hooke's Law: F 1 =k⋅x 1 20 gw=k⋅4 cm k= 4 cm 20 gw =5 gw/cm 2. Calculate the weight of the toy bear: Full length with the toy bear: L 2 =18 cm Extension (x 2 ) with the toy bear: x 2 =L 2 −L 0 =18 cm−10 cm=8 cm Now, use Hooke's Law again with the calculated spring constant k and the extension caused by the toy bear (x 2 ) to find the force (weight) of the toy bear (F 2 ): F 2 =k⋅x 2 F 2 =5 gw/cm⋅8 cm F 2 =40 gw 3. Check against the elastic limit: The elastic limit is 100 gw. The calculated weight of the toy bear is 40 gw, which is less than 100 gw, so the spring is still within its elastic limit. Therefore, the weight of the toy bear is 40 gw. Comparing with the options: 20gw 40gw 60gw 90gw The correct answer is The final answer is B .