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A heavy chain with a mass per unit length ? is pulled by the constant force F along a horizontal surface consisting of a smooth section and a rough section.The chain is initially at rest on the rough surface with x=0 .If the coefficient of kinetic friction between the chain and rough surface is µ , determine the velocity of the chain when x=L . I am applying work energy theorem . Work done by constant Force will be Force × displacement of centre of mass i.e FL but not able to find work done by friction .The friction force at an instant when chain length x lies on the rough surface should be µ?xg.This force is continuously decreasing .i feel calculus is involved here but i am unable to apply it.Please help me .thanks.

A heavy chain with a mass per unit length ? is pulled by the constant force F along a horizontal surface consisting of a smooth section and a rough section.The chain is initially at rest on the rough surface with x=0 .If the coefficient of kinetic friction between the chain and rough surface is µ , determine the velocity of the chain when x=L .

I am applying work energy theorem . Work done by constant Force will be Force × displacement of centre of mass i.e FL but not able to find work done by friction .The friction force at an instant when chain length x lies on the rough surface should be µ?xg.This force is continuously decreasing .i feel calculus is involved here but i am unable to apply it.Please help me .thanks.

Grade:12

1 Answers

Vishal
12 Points
5 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−−−−−−−−−√

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