The Missing Function

Classical Mechanics Level pending

Imagine there is a spring with equillibrium length L L , spring constant k k and mass m m uniformly distributed floating undeformed such that its upper limit is at a height H H , and then is dropped. When the spring hits the ground, it will displace from its equillibrium length a distance l l , before moving upwards. The value of H H for which l l is minimum can be written as: L − m g α k L-\frac{mg}{\alpha k} Find the value of α \alpha Click here for more problems


The answer is 8.

Hjalmar Orellana Soto by Hjalmar Orellana Soto , 24, Chile

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1 solution

Initially the energy of the system is purely gravitational: E 0 = m g ( H − L 2 ) E_{0}=mg(H-\frac{L}{2}) At the moment the spring is displaced the distance l l there is gravitational and elastic energy: E f = m g ( L − l 2 ) + k 2 l 2 E_{f}=mg(\frac{L-l}{2})+\frac{k}{2}l^2 As there are only conservative forces acting, energy is conserved: E 0 = E f E_{0}=E_{f} ⇒ m g ( H − L 2 ) = m g ( L − l 2 ) + k 2 l 2 ⇒ k l 2 − m g l + 2 m g ( L − H ) = 0 \Rightarrow mg(H-\frac{L}{2})=mg(\frac{L-l}{2})+\frac{k}{2}l^2 \Rightarrow kl^2-mgl+2mg(L-H)=0 ⇒ l = m g + ( m g ) 2 − 8 m g k ( L − H ) 2 k \Rightarrow l=\frac{mg+\sqrt{(mg)^2-8mgk(L-H)}}{2k} So, the minimum value of l l is reached when: ( m g ) 2 − 8 m g k ( L − H ) = 0 ⇒ H = L − m g 8 k (mg)^2-8mgk(L-H)=0 \Rightarrow H=L-\frac{mg}{8k} So α = 8 \alpha=8

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