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A force of `20 N` is applied on upper block as shown in the figure. Te total work done by frictional during the time interval in which the upper block has a displacement of `15 m` with respect to the ground is (take `g = 10 m//s^(2)` ) A. `50 J`B. `-50 J`C. `75 J`D. `- 75 J` |
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Answer» Correct Answer - B Block and plank both slide so, `a_("Block") = (20 -5)/(2) = (15)/(2)` `a_("plank") = (5)/(2)` `S = ut + (1)/(2)at^(2)` `15 = 0 + (1)/(2) (15)/(2)t^(2)` `rArr t = 2 sec` Desplacement of plane `= 0 + (1)/(2) (5)/(2)2^(2)` So, displacement of block with respect to plank `= 15 - 5 = 10m` `W_(f) = - 10 xx 20 xx 1//4 = -50J` |
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