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A body is desplaced from origin to `(2m, 4m)` under the following two forces: (a) `F=(2^hati + 6^hatj)N`, a constant force (b) `F(2x^hati + 3y^(2)hatj)N` Find word done by the given forces in both cases. |
Answer» Correct Answer - B (a) `F=(2hati + 6hatj)N` `dr=(dxhati + dyhatj)` `:. fdr= 2 dx + 6 dy` `W =int_((0.0))^((2m.4m))F.dr = int_((0.0))^((2m.4m))(2dx + 6dy)` `=[2x + 6y]_(0,0)^(2m,4m) =(2xx2 + 6xx4)` `=28J` |
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