A metre stick weighing 600 g, is displaced through an angle of 60° in vertical plane as shown. The change in its potential energy is (g = 10 ms?) (a) 1.5 J (b) 15 ) 6) 30 J (d) 45 J
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Answer:
Correct option is
A
1.5 J
Given,
m=500gm=0.6kg
θ=60
0
Change in potential energy at the center of the meter stick,
ΔU=U
f
−U
i
ΔU=(−mghcosθ)−(−mgh)=mgh−mghcos60
0
ΔU=mgh(1−
2
1
)=
2
mgh
ΔU=
2
0.6×10×
2
1
ΔU=1.5J
Explanation:
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