Respuesta :
Answer:
Explanation:
the putty (W[tex]_{p}[/tex] = 2.3lb) strikes the block with velocity ( here h = 8.7ft):
V[tex]_{p}[/tex] = [tex]\sqrt{2gh} = \sqrt{2.3(32.2)8.7} = 25.383ft/sec[/tex]
conservation of linear momentum gives (where W[tex]_{b}[/tex] = 22lb):
[tex]m_{p}v_p = (m_{p}+m_{b})v^{l}[/tex]
[tex]v^{i} = \frac{m_{p}v_{p}}{m_{p} + m_{b}} =\frac{W_{p}v_{p}}{W_{p}+W_{b}} = \frac{2.3(25.383)}{2+22} =2.432ft/sec[/tex]
since k = 4lb/in and there are two springs, the initial spring deflection is
δ[tex]_{0}[/tex] = [tex]\frac{W_{b}}{2k} = \frac{22}{2.4}[/tex] = 2.75in
conservation of energy gives (datum is the initial position of the block and we divide with 12 to get deflection in inches):
ΔT + ΔV[tex]_{g}[/tex] + ΔV[tex]_{e}[/tex] = 0
[tex]0 - \frac{1}{2}(m_{p}+m_{b})v^{'2} +(0-m_{p}+m_{b})gd+\frac{1}{2}m_{p}k((d_{0}+d)^{2} - d\frac{2}{0} ) = 0[/tex]
[tex]\frac{1}{2} \frac{2.3+22}{32.2} (2.432)^{2} -(2+22)\frac{d}{12} +\frac{1}{2}2.3.4((2.75 + d)^{2}-2.75^{2} ) =0[/tex]
[tex]-2.203-4.216d+\frac{12.65+11+d^{2}+7.56 }{4.6} =0[/tex]
