InterviewSolution
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InterviewSolution
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Two blocks of equal mass `m` are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force `F` is applied on the first block pulling away from the other as shown in Fig. If the extension of the spring is `x_(0)` at time `t`, then the displacement of the second block at this instant isA. `(Ft^(2))/(2m)-x_(0)`B. `1/2((Ft^(2))/(2m)+x_(0))`C. `1/2((2F^(2))/m-x_(0))`D. `1/2((Ft^(2))/(2m)-x_(0))` |
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Answer» Correct Answer - D Suppose the displacement of the first block is `x_(1)` and that of the second is `x_(2)` then `x=(mx_(1)+mx_(2))/(2m), (Ft^(2))/(4m)=(x_(1)+x_(2))/2` or `x_(1)+x_(2)=(Ft^(2))/(2m)`……….i Further the extension of the spring is `x_(1)-x_(2)`. Therefore `x_(1)-x_(2)=x_(0)`.....ii From eqn i and ii `x_(1)=1/2((Ft^(2))/(2m)+x_(0))impliesx_(2)=1/2((Ft^(2))/(2m)-x_(0))` |
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