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Figure shows (x,t) (y,t) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 g , find the force (direction and magnitude) acting on the particle . |
Answer» Clearly from diagram (a) , the variation can be related as x = t `implies (dx)/(dt) = 1` m/s `a_(x) = 0` From diagram (b) `" " y = t^(2)` `implies " " (dy)/(dt) = 2t` or `a_(y) = (d^(2)y)/(dt^(2)) = 2 m//s^(2)` Hence , `" " I_(y) = ma_(y) = 500 xx 10^(-3) xx 2 = 1N " " (because m = 500 g)` `F_(x) = ma_(x) = 0` Hence , net force , `" " F = sqrt(F_(x)^(2) + F_(y)^(2)) = F_(y) = 1N " " `(along y-axis) |
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