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A body is moving down a long inclined plane in inclination `45^(@)` with horizontal. The coefficient of friction between the body and the plane varies as `mu=x//2`, where `x` is the distance moved down the plane. Initially `x=0 & v=0`.A. When `x=2` the velocity of the body is `sqrt(gsqrt(2))m//s`B. The velocity of the body increases all the timeC. At an instant when `v!=0` the instantaneous acceleration of the body down the plane is `(g(2-x))/(2sqrt(2))`D. The body first accelerates and then decelerates |
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Answer» Correct Answer - A::C::D When `v cancel=0` the aceeleration is `vec(a)_("inst") = (1)/(m)(mg sin theta-mu mg cos theta)` `= g(sin theta-(x)/(2) cos theta)-(g(2-x))/(2sqrt(2))` `rArr v(dv)/(dx) = (g)/(2sqrt(2))(2-x)` `rArr underset(0)overset(v)int v dv = (g)/(2sqrt(2))[2x-(x^(2))/(2)]_(0)^(2) rArr v = sqrt(gsqrt(2))m//s` Also it is clear that for `x lt 2` the body accelerates, at `x = 2` the acceleration is zero and for `x gt 2` the body retards till it comes to rest. |
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