Part A: If the mass leaves the end of the spring at a speed of 4.43 class=\"math
ID: 2123941 • Letter: P
Question
Part A: If the mass leaves the end of the spring at a speed of 4.43class="mathjax_preview">m/sstyle="border-left: 0em solid; display: inline-block; overflow: hidden; width: 0px; height: 1.404em; vertical-align: -0.389em;">style="display: inline-block; width: 0px; height: 2.154em;">style="font-family: mathjax_main;" id="mathjax-span-40" class="mi">style="font-family: mathjax_main;" id="mathjax-span-39" class="mo">id="mathjax-span-38" class="mrow">id="mathjax-span-37" class="texatom">style="font-family: mathjax_main;" id="mathjax-span-36" class="mi">id="mathjax-span-35" class="mrow">id="mathjax-span-34" class="texatom">id="mathjax-span-33" class="mrow">style="position: absolute; clip: rect(1.219em, 1000em, 2.588em, -0.529em); top: -2.154em; left: 0em;">style="display: inline-block; position: relative; width: 27px; height: 0px; font-size: 125%;">id="mathjax-span-32" class="math"> , what is the spring constant of the spring?style="" role="textbox" id="mathjax-element-7-frame" class="mathjax">class="type-specific" type="tex">id="yui_3_3_0_23_137615518422073" class="replace-insert done">
Explanation / Answer
U1 = 1/2kx^2
KE 2 = 1/2mv^2
PE2 = mg (0.283)*sin 20 = 4.951 J
U 1= KE2 +PE2
kx^2 = mv^2 + (2*4.951)
k =( mv^2 + 4.951)/x^2
=((5.22*(4.43)^2) +4.951)/ .283^2
spring constant = 1402.745 N/m
2)
Initial Potential energy = 0.5*K*x^2 = 0.5*1402.745*0.283^2 =56.172 J
final P.E = mgh = mg*L sin 20 = 5.22*9.8*L * sin 20 = 17.496 L
Final K.E = 0.5*m*V'^2 = 0.5*5.22*2.81^2 = 20.608 J
so, P.E1 +K.E1 = P.E2 +K.E2
0 +56.172 = 20.608 +17.496 L
solving L = 2.033 m