Vishal
Last Activity: 7 Years ago
The chain is initially at rest, soKEo=0KEo=0The force of friction is given byf(x)=μρ(L−x)gf(x)=μρ(L−x)gThe net force on the chain is∑F=F−μρ(L−x)g∑F=F−μρ(L−x)gWork done on the chain is the integral of force over distance, soW=∫L0F−μρ(L−x)gdxW=∫0LF−μρ(L−x)gdxIntegrate and getW=FL−12μρgL2W=FL−12μρgL2Use Work-Energy TheoremKEf=KEo+WKEf=KEo+Wand final kinetic energy isKEf=FL−12μρgL2KEf=FL−12μρgL2Kinetic Energy equationKEf=12mv2f=12(ρL)v2f=FL−12μρgL2KEf=12mvf2=12(ρL)vf2=FL−12μρgL2Solve for final velocityvf=2Fρ−μgL−−−−−−−−−√