First slide

Work Energy Theorem

Question

A long rod ABC of mass “m” and length “L” has two particles of masses “m” and “2m” attached to it as shown in the figure. The system is initially in the horizontal position. The work to be done to keep it vertical with A at the bottom is (g = acceleration due to gravity)

Easy

A

2mgl
B

3mgl
C

5mgl/2
D

3mgl/2

Solution

$\large w = mg\frac{L}{2} + mg\frac{L}{2} + (2m)gL = 3mgL$

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