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A block is kept on a rough horizontal surface. A variable horizontal force is acting on the block as shown in the figure. At `t=0`, block is at rest. Kinectic energy of block at `t=5 sec` will be `(g=10m//s^(2))` A. 60 JB. 160 JC. 80 JD. zero |
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Answer» Correct Answer - C `(f_(s))_(max) = 30 N,f_(x) = 20 N` block will start moving at `t=3 sec` after `3 sec` by `F_("net") = ma` `rArr 10 t - 20 = 10 a rArr a =t -2` `underset(0)overset(v)(int) dv = underset(3)overset(5)(int) (t-3)dt=v=4 m//s` `rArr K.E = (1)/(2) xx 10 xx 4_(2) = 80 J` |
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