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Step-by-Step Solution
Step 1: Identify the mass of the hanging part
The chain is uniform, with total length 2 m and total mass 4 kg. A length of 0.60 m is hanging. Hence, the fraction of the chain hanging is
$ \frac{0.60}{2} = 0.30 $ (i.e., 30% of the length).
So, the mass of the hanging part $m'$ is:
$ m' = 4 \times \frac{0.60}{2} = 1.2\ \text{kg} $.
Step 2: Determine the initial vertical position of the center of mass of the hanging part
Since 0.60 m of the chain hangs from the table, its center of mass will be exactly at half of this length from the table’s edge, which is:
$ x = \frac{0.60}{2} = 0.30\ \text{m} $ below the table surface.
Step 3: Choose the reference for potential energy
We set the potential energy $PE=0$ at the level of the table surface. Initially, the hanging portion’s center of mass is 0.30 m below this level.
Step 4: Calculate the initial potential energy of the hanging part
The potential energy of a mass $m'$ at a distance $x$ below the reference is:
$ U_i = -\,m' g\,x $
Substituting $m' = 1.2\ \text{kg}$, $g = 10\ \text{m/s}^2$ (assuming this simplified value from the solution), and $x = 0.30\ \text{m}$, we get:
$ U_i = -(1.2)\,(10)\,(0.30) = -3.6\ \text{J}$.
Step 5: Calculate the final potential energy of the hanging part
When the chain is fully pulled up onto the table, the entire mass is at the reference level ($PE=0$). Thus,
$ U_f = 0\ \text{J}.$
Step 6: Compute the work done in pulling the chain
Work done is equal to the change in potential energy:
$ W = \Delta U = U_f - U_i = 0 - (-3.6)\ = 3.6\ \text{J}.$
Therefore, the work done in pulling the entire chain onto the table is $3.6\ \text{J}.$
