Question
Question: A uniform chain of mass M and length L is held on a horizontal frictionless table with \[\dfrac{1}{\...
A uniform chain of mass M and length L is held on a horizontal frictionless table with π1th of its length hanging over the edge of the table. The work done in pulling the chain up on the table is
A. πMgL
B. 2πMgL
C. π2MgL
D. 2π2MgL
Solution
Consider the small element of length of the hanging portion of the chain. Derive the expression for the small work required to pull that element of the chain and integrate the equation to determine the work done to pull the whole hanging portion of the chain back on the table.
Formula used:
W=mgx
Here, m is the mass , g is the acceleration due gravity and x is the distance.
Complete step by step answer:
Let the small length dx of the hanging chain at a distance x from the table as shown in the figure below.
We can express the mass of the element dx as follows,
dm=(LM)dx
Now, we calculate the small work done by pulling the mass dm on the table as follows,
dW=(LMdx)gx
⇒dW=(LMgx)dx
Here, g is the acceleration due to gravity.
Now, we have to calculate the total work done to pull πL length of the chain back on the table as follows,
W=0∫πL(LMgx)dx
⇒W=LMg0∫πLxdx
⇒W=LMg(2x2)0πL
⇒W=LMg2(πL)2−0
∴W=2π2MgL
So, the correct answer is “Option D”.
Additional Information:
the work done by the gravitational force on the body of mass m to move a body at a distance x in the upward direction is can be expressed as,
W=mgx
The S.I. unit of work done is joule
Note:
The mass of the small element dx in the above figure is expressed as, dm=(LM)dx. Here, LM is known as linear mass density of the chain. This expression is valid only for objects which have uniform density that is the mass of the chain should not be changed with distance.