7-53 할리데이 11판 솔루션 일반물리학

(풀이자주:풀이에 테일러급수가 필요합니다. 관련한 내용은 별도의 링크로 분리했습니다. 테일러 급수 자체에 대한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.)

https://solutionpia.tistory.com/624 

테일러 급수

$$\put \exp(x)=\e^x,$$ $$\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!},~~(\because \text{by Taylor Series})$$ $$\therefore \exp(-kx^2)= \sum_{n=0}^{\infty}\frac{(-kx^2)^n}{n!} $$ $$ \begin{cases} x_i&=0.60\ut{m}\\ x_f&=2.10\ut{m}\\ F&=\exp(-5x^2)\\ \end{cases} $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W&=\int_{i}^{f} \exp(-kx^2) \cdot \dd x\\ &=\int_{i}^{f} \sum_{n=0}^{\infty}\frac{(-kx^2)^n}{n!} \cdot \dd x\\ &=\sum_{n=0}^{\infty}\bra{\frac{(-k)^n}{n!}\int_{i}^{f} x^{2n} \cdot \dd x}\\ &=\sum_{n=0}^{\infty}\frac{(-k)^n ({x_f}^{2 n+1}-{x_i}^{2 n+1})}{(2 n+1) n!}\\ &=\sum_{n=0}^{\infty}\frac{(-5)^n (0.6^{2 n+1}-2.1^{2 n+1})}{(2 n+1) n!}\\ \end{aligned} $$ $$\put W(t)=\sum_{n=0}^{t}\frac{(-5)^n (0.6^{2 n+1}-2.1^{2 n+1})}{(2 n+1) n!},$$ $$ \begin{aligned} W(0)&=-1.5\\ W(10)&\approx -4.9357774121747375\times10^5\\ W(20)&\approx -7.864220191714263\times10^6\\ W(30)&\approx -1.0696534196060577\times10^6\\ W(40)&\approx -5.991942239765855\times10^3\\ W(50)&\approx -3.0485422420840154\\ W(60)&\approx -2.3120711378363985\times10^{-2}\\ W(70)&\approx -2.2899938329124892\times10^{-2}\\ W(80)&\approx -2.289993513545718\times10^{-2}\\ W(90)&\approx -2.2899935135445686\times10^{-2}\\ W(100)&\approx -2.2899935135445686\times10^{-2}\\ &\vdots \end{aligned} $$ $$ \begin{aligned} \therefore W&\approx -2.28999..\times10^{-2}\ut{J}\\ &\approx -2.29\times10^{-2}\ut{J}\\ &\approx -22.9\ut{mJ}\\ \end{aligned} $$