A 0.2 kg ball is suspended from a thread of 1m. it is pulled at an angle o 30 degree with the vertical. calculate the work done against gravity
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Here,
θ = 30°
L = 1 m
m = 0.2 kg
g = 10 ms-2
Work done is stored as PE in the bob.
PE = mgh
= 0.2×10×1×(1 – cos300)
= 0.2×10×(1 – 0.87)
= 0.26 J
θ = 30°
L = 1 m
m = 0.2 kg
g = 10 ms-2
Work done is stored as PE in the bob.
PE = mgh
= 0.2×10×1×(1 – cos300)
= 0.2×10×(1 – 0.87)
= 0.26 J
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