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题:轻质弹簧竖直放在水平面上,一质量为m的木块放到弹簧上以后,弹簧被压缩x_0。求弹簧的倔强系数K。正确的解法是:当弹簧被压缩x_0时,木块所受弹力与重力平衡,所以有mg=Kx_0即K=mg/x_0。但有的同学从机械能守恒定律去求解时却得到了另一个答案。他们的解法是:设图1(b)木板所在位置是重力势能的零势点,从木板放到弹簧上到图1(b)这一过程中重力势能转化为弹性势能,所以有:mgx_0=1/2Kx_0~2 K=2mg/x_0。与第一种解答不一致。第二种解法似乎也有一定道理,那么究竟错误在
Problem: The lightweight spring is placed vertically on a horizontal surface. After a mass m of wood is placed on the spring, the spring is compressed x_0. Find the stubborn coefficient K of the spring. The correct solution is: When the spring is compressed x_0, the elastic force of the block is balanced with gravity, so there is mg = Kx_0 that is K = mg/x_0. However, some students got another answer when they solved the law of conservation of mechanical energy. Their solution is: Let Fig. 1(b) the position of the wood board be the zero potential point of the gravitational potential energy. From the board to the spring, the gravitational potential energy is converted into elastic potential energy in the process of Fig. 1(b), so there is: mgx_0= 1/2Kx_0~2 K=2mg/x_0. Contrary to the first solution. The second solution seems to have some truth, so what’s wrong is