Answer to Question #96571 in Mechanics | Relativity for Ebube Paul Nwokeji

Question #96571
A Glider with mass equals 0.2kg sits on a frictionless horizontal air track, connected to a spring of negligenble mass with force constant K=5.0N/M. You pull on the Glider stretching the spring 0.100m and then release it with no initial velocity. The Glider begins to move back towards its equilibrium position(S=0).
A) What is its speed when x=0.0800
B) What is the value of x when the object speed is 0.15m/s?
1
Expert's answer
2019-10-16T09:30:46-0400

A) From the conservation of energy:


0.5mv2+0.5kx2=0.5kA20.5mv^2+0.5kx^2=0.5kA^2

v=km(A2x2)v=\sqrt{\frac{k}{m}(A^2-x^2)}

v=50.2(0.120.082)=0.3msv=\sqrt{\frac{5}{0.2}(0.1^2-0.08^2)}=0.3\frac{m}{s}

B) From the conservation of energy:

0.5mv2+0.5kx2=0.5kA20.5mv'^2+0.5kx'^2=0.5kA^2

x=A2mkv2x'=\sqrt{A^2-\frac{m}{k}v'^2}

x=0.120.250.152=0.095 mx'=\sqrt{0.1^2-\frac{0.2}{5}0.15^2}=0.095\ m


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