Question

A heavy chain with mass per unit length ‘ρ’ 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 the rough surface is µk, then what is the velocity v (in m/s) of the chain when x = L, if the force F is greater than µk ρgL in order to initiate the motion. : If F = 21N, μ= 0.5, L = 1 m, ρ = 2 kg/m

Answer 1

Answer:

Using work energy theorem

Work done by F+work done by friction=change in kinetic energy of chain

FL−∫  

0

L

​  

μ  

k

​  

N(x)dx=  

2

1

​  

mv  

2

...(I)

N(x) reaction force of the rough region which is equal to weight of chain in rough region

N(x)=P(L−x)g

m is weight of total chain

m=PL

Substituting values of m and N in equation (I)

FL−  

2

1

​  

μ  

k

​  

PL  

2

g=  

2

1

​  

PLv  

2

 

v=  

P

2F

​  

−μ  

k

​  

Lg

​

Explanation:

Answer 2

Answer:

A heavy chain with mass per unit length ‘ρ’ is pulled by the constant force etween the chain and the rough surface is µk, then what is the velocity v (in m/s) of the chain when x = L, if the force F is greater than µk ρgL in order to initiate the motion. : If F = 21N, μ= 0.5, L = 1 m, ρ = 2 kg/m