Q:A 5.1kg block is pulled along a frictionless floor by a cord that exert a force P = 10N at angle thyta = 37 degree , above the horizontal ,
- The force P is slowly increased . what is the value of P just before the block breaks off the floor.
- What is the acceleration of the block just before it is lifted off the floor.
Please find below the solution to the asked query:
To lift the block, the vertical component of the applied force should be equal to its weight. Therefore,
(ii) Now, Just before the block leaves the contact force, then the applied force n horizontal direction is.
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