JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 23)
As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s2)
The value of 'x' to the nearest integer is __________.
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The value of 'x' to the nearest integer is __________.
Answer
10
Explanation
By energy conservation
Ki + Ui = Kf + Uf
0 + 10 $$\times$$ 10 $$\times$$ 10 = $${1 \over 2}$$ $$\times$$ 10 $$\times$$ v$$_B^2$$ + 10 $$\times$$ 10 $$\times$$ 5
1000 = 5v$$_B^2$$ + 500
v$$_B^2$$ = $${{500} \over 5}$$ = 100
VB = 10 m/s
x = 10
Ki + Ui = Kf + Uf
0 + 10 $$\times$$ 10 $$\times$$ 10 = $${1 \over 2}$$ $$\times$$ 10 $$\times$$ v$$_B^2$$ + 10 $$\times$$ 10 $$\times$$ 5
1000 = 5v$$_B^2$$ + 500
v$$_B^2$$ = $${{500} \over 5}$$ = 100
VB = 10 m/s
x = 10
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