最大赢亏功为 C 5)J-∽450C(+9) Al max n2[l] 1)W=M2r=W=3140 M=3140 ≈500Nm) W=J。Md=zM,221x2=zM 3140 M =1000(Nm) 2)画出M-④图,如下图所示。设a动能为E,则 000 E=E+△W=E+abM4=E。+。500 E=E+AW,=En--bcM,=Eo--500 E4=E+△W3=E+=cM4=E+,500 16 e=E+AW=E--deM=e--500 E=E+Als=E+efM,=E,+500 ER=Er E,+-fgM,=E En=E2+△W7=E E E-E=E-E=-x500=125(N.m)3 最大赢亏功为 max 2 C  = W  5） max 2 2 2 1 1 450 ( ) [ ] [ ] 9 f e m W C J J J J     n  = − = − + 4、 1） 2 3140 W M W d d r = = =  3140 500(N m) 2 Md  =   2 max max max 0 1 1 2 2 2 2 W M d M M M r r r r r   = = +  =     max 3140 1000(N m) r r W M   = = =  2）画出 M − 图，如下图所示。设 a 动能为 Ea ，则 a) b) 1 1 500 2 8 E E W E abM E b a a d a  = +  = + = + 2 1 500 2 8 E E W E bcM E c b b d a  = +  = − = − 3 1 500 2 16 E E W E cdM E d c c d a  = +  = + = + 4 1 500 2 16 E E W E deM E e d d d a  = +  = − = − 5 1 500 2 16 E E W E ef M E f e e d a  = +  = + = + 6 1 500 2 16 E E W E fgM E g f a d a  = +  = + = − ' 7 1 ' 2 E E W E ga M E a g a d a = +  = + = max max min 1 500 125(N m) 4  = − = − = =  W E E E E b c 