A 3.0 m long steel chain is stretched out along the top level of ahorizontal sca

Question

A 3.0 m long steel chain is stretched out along the top level of ahorizontal scaffold at a construction site, in such a way thatx = 1.5 m of the chain remains onthe top level and y = 1.5 m hangsvertically. (See the figure.) At this point, the force on thehanging segment is sufficient to pull the entire chain over theedge. Once the chain is moving, the kinetic friction is so smallthat it can be neglected. How much work is performed on the chainby the force of gravity as the chain falls from the point where1.5 m remains on the scaffold to the pointwhere the entire chain has left the scaffold? (Assume that thechain has a linear weight density of 30N/m.) 1 J

Explanation / Answer

   Work done by gravity = loss of potential energy =(mgh)    Assume the lower end at initial position of chainas h=0 As we consider the height of center of mass of the topand overhanging portions,   initial gravitational potential energy =(m/2)*g*(y/2)(at the side) + (m/2)*g*y (on top) final PE = m*g*(h=0) = 0 So, work done = mgy/4 + mgy/2 = 3mg/4 * 1.5 =4.5/4 * 30*3 J = 101.25 J

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