Question

# In Fig. $$8-45,$$ a chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length $$L = 28 \,cm$$ and mass $$m = 0.012\, kg,$$ how much work is required to pull the hanging part back onto the table?

Solution

## The work required is the change in the gravitational potential energy as a result of the chain being pulled onto the table. Dividing the hanging chain into a large number of infinitesimal segments, each of length $$dy$$, we note that the mass of a segment is $$(m/L) dy$$ and the change in potential energy of a segment when it is a distance $$|y|$$ below the tabletop is                    $$dU = (m/L) g|y| dy = -(m / L) gy\,dy$$since y is negative-valued (we have $$+y upward and the origin is at the tabletop). The total potential energy change is$$\Delta U = -\dfrac{mg}{L} \int_{-L/4}^0 \,y \,dy = \dfrac{1}{2} \dfrac{mg}{L} (L / 4)^2 = mgL / 32.$$The work required to pull the chain onto the table is therefore$$W = \Delta U = mgL / 32 = (0.012 \,kg)(9.8\,m/s^2)(0.28\,m) / 32 = 0.0010 \,J.Physics

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