InterviewSolution
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InterviewSolution
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A horizontal plane supports a plank with a block placed on it. A light elastic string is attached to the block, which is attached to a fixed point O. Initially, the cord is unstretched and vertical. The plank is slowly shifted to right until the block starts sliding over it. It occurs at the moment when the cord deviates from vertical by an angle `theta=0^@`. Work done by the force F equals A. (a) Energy lost against friction `F_1` plus strain energy in cordB. (b) Work done against total friction acting on the plank aloneC. (c) Work done against total friction acting on the plank plus strain energy in the cordD. (d) Work done against total friction acting on the plank plus strain energy in the cord minus work done by friction acting on the block |
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Answer» Correct Answer - A::B::D When the plank is shifted, the cord elongates and becomes inclined with vertical. Therefore, a tenstion is developed in the cord and that tension has two components, a vertical component and a horizontal component: Horizontal component tries to slide the block leftwards, relative to the plank. But friction `F_2` prevents that slipping. Hence, the block moves to the right with the plank. Since the block does not slip over the plank, no energy is lost against friction `F_2`. Hence, work done by the force F is used to overcome loss against friction `F_1` and to elongate the cord. Therefore, option (a) is correct. The cord elongates due to displacement of the block and the block gets displaced due to friction `F_2` acting on it. Hence, the friction `F_2` is responsible for the elongation of the cord. Therefore, strain energy stored in the cord is equal to work done by the friction on the block or work done against friction acting on the upper surface of the plank. Hence, the total work done by F is equal to energy lost against friction `F_1` plus work done against friction acting on upper surface of the plank. Therefore, option (b) is correct. Hence, it is obvious that option (c) is wrong. Since, strain energy stored in the cord is equal to work done by friction acting on the block, option (d) becomes same as option (b). Hence, it is also correct. |
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