The origin of the coordinate system is established at the fixed point of the spring, with x as the integral variable, and the integral interval is [0.25, 0.4]. At this time, at any position x in the interval [0.25, 0.4], the spring elongation is ( X-0.25), then W = 0.40.25 (x - 0.25) 985 dx = 0.225 (J). Second: the origin of the coordinate system is established at the equilibrium position of the spring. When the spring is extended from 25cm to 40cm, the spring is extended by the equilibrium position by 0.4-0.25=0.15 (m). Taking x as the integral variable, the integral interval is [0, 0.15]. At this time, at any position x in this interval, the spring elongation is (x-0)=x, then there is W=0.150. 985xdx=0.2205(J) It can be seen from the two solutions of this problem that the first method considers the original length of the spring, and the elongation at the position x of the spring is 'x=x-spring length. Method 2 does not consider the elongation of the spring at the position x of the spring (x=x-0=x. Look at the second example. When the length of the metal rod is extended from a to a+x, the required force is required. For kax (k is a constant), try to work on the metal rod from the length a to b. If the elastic coefficient is K, then K(a+x-a)=kax, and K=kax, so method 1: do not consider the length of the rod, when a is extended to b, the rod is extended (b-a), The integration interval is [0, b-a], and f=kax at this time, so w=b-a0kaxdx=k(b-a)22a(J) Method 2: Consider the rod length, and set the integration interval to [a,b] When the rod is elongated to x (x∈, [a, b]), the rod is elongated (x-a), and f=ka(x-a), so w=baka(x-a) Dx=k(b-a)22a(J) The above two questions are elongated. Let's look at the compression situation. For example, the original length is 0.3 meters, and each compression of 0.01 meters requires 2N, and the work of compressing the spring from 0.25 meters to 0.2 meters is sought. The solution takes the fixed point of the spring as the origin, and the integral interval is [0.2, 0.25], then the spring is compressed (0.3-x), ie f=(0.3-x)k, where k=2/0.01=200, then w=0.250.2 200(0.3-x)dx=0.75(J) Solution 2 takes the equilibrium position of the spring as the origin. When the spring is compressed from 0.25 m to 0.2 m, the springs are compressed by 0.05 and 0.1 respectively, so the integral interval is [0.05. 0.1]. For any x ∈ [0.05, 0.1], x is the compression of the spring, so there is f = 200x, that is, w = 0.10.05. Solar Landscape Light,Solar light.outdoor solar garden lamp Solar Landscape Light,Solar light.outdoor solar garden lamp,solar light Shenzhen You&My Electronic Technology Co., Ltd , https://www.ymledtrade.com