InterviewSolution
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InterviewSolution
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A force of F=0.5 N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, g=10 `ms^(-2)`) A. `-0.5 J`B. 0.5 JC. 2 JD. `-2 J` |
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Answer» Correct Answer - B (b) Maximum acceleration of 1 kg block may be `a_(max)=mug=1ms^(-2)` Common acceleration, without relative motion between two blocks may be, `a=(0.5)/(3)ms^(-2)` Since, `a lt a_(max)` There will be no relative motion and blocks will move with acceleration `(0.5)/(3)ms^(-2)`. Force of friction by lower block on upper block, `f=ma=(1)((0.5)/(3))=(1)/(6)N("towards right")` `:. " " W=fxxs=0.5J` |
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