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In the figure shown upper block is given a velocity of `6 m//s` and lower block `3 m//s`. When relative motion between them is stopped. .A. Work done by friction on upper block is negativeB. Work done by friction on both blocks is positiveC. The magnitude of work done by friction on upper block is `10 J`D. Net work done by friction is zero. |
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Answer» Correct Answer - A::C From conservation of linear momentum `(1+2) v = (6 xx 1)+(2-3)` `:. v = 4m//s` (of both the blocks) from work energy theorm i.e., `W_("total") = Delta KE` on `1 kg` block, `W_(f) = (1)/(2) xx 1 xx (4^(2) - 6^(2)) = - 10 J` on `2 kg` block `W_(f) = (1)/(2) xx 2(4^(2) - 3^(2)) = + 7 J` `:.` Net work done by friction is ` - 3 J`. |
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