Kindly solve wpe cpp no11

Kindly solve wpe cpp no11 11. 12. One meter stick having mass 600 gm, is pivoted at one end and is displaced through an angle of 600, the increase in its potential energy is (g = 10 rn/s2) (A) 1.5 J (C) 150 J (B) 15 J (D) 0.15 J Two particles 1 and 2 are allowed to descend on two frictionless chords OP and OQ starting from O. The ratio of the speeds of the particles 1 and o 1

Dear student,
The length of the stick is 1 metre so its centre of mass is at 0.5 metre from either end.
Now, on displacement of 60⁰ of the stick the COM would rise -----
0.5 - 0.5cos60⁰ = 0.5 x (1 - 0.5) = 0.25metre
Thus new location of COM is 0.25 m above the old location. and therefore the increase in PE would be,
0.6 kg x 10ms^-2 x 0.25 m = 1.5 Joules

Regards

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