InterviewSolution
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InterviewSolution
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A mat of mass 1kg and length 1m is placed on the floor. One end of the mat is pulled with a constant speed of 1m/s towards the other end till the other end comes in to motion (till the mat is reverse). How much force is required to do this ? |
Answer» Solution : As shown in figure, a mat is being pulled with a coonstant speed of `v = 1m//s,` so that the mass of the part of the mat is continuously increasing. Hence here the mass is a variable. The time required for bringing the entire mat in motion is given by `Delta t =("DISTANCE covered by the end")/("speed")` `= (2m)/(1m//s) =2s` (Distance covered by the end =1m + 1m =2m) From Newton.s second law of motion, `F _(n et) = (Delta p)/(Delta t) = (Delta (mv))/(Delta t)` Here v is constnat, so we get `F _( n et) =v (Deltam)/(Delta t)` Where `Deltam` is the cahrge of mass in `Delta t` time. THge charge of mass in 2s is equal to entire mass of mat. ` F _(n et) = ((1m //s) X (1kg))/(2s)` `=1/2N` In the horizontal direction only ONE FORCE is acting. Hence the required force is `1//2N` |
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