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work done by a variable force

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work done by a variable force
In a one-dimensional situation,a force given by F(x) = (2.2 − 0.61x2) N acts over a displacement from x = 0 m to x = 0.12 m.What work does the force perform?
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A spring,compressed 0.31 m from its equilibrium position,is released and returns to its equilibrium position.In doing so it pushes a projectile and accelerates it.The spring constant is 75 N/m.How much work does the spring perform on the projectile?
求大神指教下这两题
我实在不会做了
work done by a variable force
在一维的情况下,力F（x）=（2.2−0.61x2）N 作用于一个物体,当（物体）有从x = 0 m到x = 0.12 m的位移时,该力做了多少功?
力做功W=∫[x=0,x=0.12]F（x）dx=∫[x=0,x=0.12]（2.2−0.61x2）dx=2.2·0.12-0.61/3(0.12³-0)=0.26364864 J
答：力做功为0.26364864焦.
一个弹簧从它的平衡位置被压缩0.31米,然后被释放并返回到其平衡位置,在此过程中,弹簧将弹丸推动并加速.弹簧的常数为75 N/m,弹簧对弹丸做多少功?
弹簧对弹丸施加的力F（x）=kx,X=0.31米,弹簧做功为W=∫[0, 0.31]F（x）dx=0.5kX²=0.5·75·0.31²=0.0360375（J）
答：弹簧做功0.0360375焦.
再问： 第二题能在看下么 貌似结果不对啊
再答： 第二题用的是计算器，应该是3.60375焦。
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