Work Energy and deflection Problem

Tags:
1. Nov 9, 2015

Jonski

1. The problem statement, all variables and given/known data
The 2 kg piece of putty is dropped 2m onto the 18 kg block initially at rest on the two springs, each with a stiffness k =1.2kN/m . Calculate the additional deflection of the springs due to the impact of the putty, which adheres to the block upon contact.

2. Relevant equations
W = ΔKe + ΔEp +ΔEs

3. The attempt at a solution
I started by saying that there is no external force so W = 0.
Next I found the speed of the block just as the putty hits it.
v1 = putty velocity = v^2 = 2*9.8*2
v = 6.26m/s

P1 = P2
m1v1 = (m1+m2)v2
2*6.26 = (2+18)v2
v2 = 0.63m/s

ΔKe = 1/2m(Δv^2)
= 1/2 * 20 * (0 - 0.63^2) (max defl at v = 0)
=-3.92J

ΔEp = mgΔh
=20*9.8*δ
=196δ J

ΔEs = 1/2k(x^2 - x^2)*2
= 1200((0.0735+δ)^2-(0.0735)^)
=1200(0.147δ +δ^2)
=176.4δ+1200δ^2

0 = 1200δ^2+372.4δ-3.92
δ=0.204m

Just want to know if this is right, as I don't think it is. Also would be great if you could tell me where I went wrong. Thanks

Last edited: Nov 9, 2015
2. Nov 10, 2015

Simon Bridge

To troubleshoot: at each stage, write a short statement describing the energy change, or other process, and keep the variables... do the algebra 1st then plug the numbers in.
i.e. step 1, putty $m_1$ falls distance $h$ exchanging gpe for ke... so $v_1^2 = 2gh/m_1$ ... gaining momentum $p = \sqrt{2ghm_1}$ ... which is conserved... kinetic energy is $p^2/2m$ ...

Take care to be consistent with variable names... ie k is already used for a single spring.
This should tell you if and ehere you have made a mistake.