✔ 最佳答案
I suppose your A represents the distance CD in the diagram. The Law of Conservation of Energy always holds. It is your equation that is wrong. You have missed out the elastic potential energy stored in the rope at point C. Since at this point, the rope has already extended by a length equals to BC,
Hence, based on the Law of Conservation of Energy, the equation should be written as,
KE of boy at C + PE of boy at C + elastic PE of rope at C = elastis PE of rope at D
(1/2)mv^2 + mgA + (1/2)k(BC)^2 = (1/2)k(BD)^2 ------------ (1)
where m is the mass of the boy,
A = distance CD
g is the acceleration due to gravity
k is the force constant of the rope
using BD = BC + CD = BC + A
The term (1/2)k(BD)^2 on the right hand side of the equation can be written as,
(1/2)k(BD)^2 = (1/2)k(BC+ A)^2 = (k/2)[BC^2 + 2(BC)A + A^2] ------- (2)
At point C , mg = k(BC), hence BC = mg/k
thus, (2) becomes,
(1/2)k(DB)^2 = (k/2)[(mg/k)^2 + 2mgA/k + A^2]
Equation (1) then becomes,
(1/2)mv^2 + mgA + (1/2)k(mg/k)^2 = (k/2)[(mg/k)^2 + 2mgA/k + A^2]
i.e. (1/2)mv^2 + mgA + (mg)^2/2k = (mg)^2/2k + mgA + (1/2)k.A^2
finally, we have,
(1/2)mv^2 = (1/2)kA^2
Therefore, the term mgA (i.e. 50(10)A ) in your equation is redundant.
