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प्रश्न
Figure shows (x, t), (y, t ) diagram of a particle moving in 2-dimensions.
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 (b) |
If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.
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उत्तर
Clearly, from the diagram (a), the variation can be related as x = t
⇒ `(dx)/(dt)` = 1 m/s
ax = 0
From diagram (b) y = t2
⇒ `(dy)/(dt)` = 2t or ay = `(d^2y)/(dt^2)` = 2 m/s2
Hence, Iy = may = 500 × 10–3 × 2 = 1 N .....(∵ m = 500 g)
Fx = max = 0
Hence, net force, F = `sqrt(F_x^2 + F_y^2)` = Fy = 1 N .....(along y-axis)
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