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A body of mass m rests on a horizontal plane with the friction coefficient k. At the moment t=0 a horizontal force is applied to it, which varies with time as F=at, where a is a constant vector. Find the distance traversed by the body during the first t seconds after the force action began. |
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Answer» Solution :Since, the applied force is proportional to the TIME and the frictional force also EXISTS, the MOTION does not start just after applying the force. The body starts its motion when F equals the limiting friction. Let the motion start after time `t_0`, then `F=at_0=kmg` or, `t_0=(kmg)/(a)` So, for `t=let_0`, the body remains at rest and for `tgtt_0` obviously `(mdv)/(dt)=a(t-t_0)` or, `mdv=a(t-t_0)dt` Integrating, and noting `v=0` at `t=t_0`, we have for `tgtt_0` `UNDERSET(0)overset(v)intmdv=aunderset(t_0)overset(t)int(t-t_0)dt` or `v=(a)/(2m)(t-t_0)^2` Thus `s=int vdt=(a)/(2m)underset(t_0)overset(t)int (t-t_0)^2dt=(a)/(6m)(t-t_0)^3` |
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