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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.

Dear Student,

Please find below the solution to the asked query:

*(i)*`

To lift the block, the vertical component of the applied force should be equal to its weight. Therefore,

$P\text{'}\mathrm{sin}{37}^{0}=5.1\times 10\Rightarrow P\text{'}=\frac{5}{3}\times 51=5\times 17\phantom{\rule{0ex}{0ex}}\Rightarrow P\text{'}=85N$

*(ii)* Now, Just before the block leaves the contact force, then the applied force n horizontal direction is.

$ma=P\mathrm{cos}\theta \Rightarrow a=\frac{85\times \mathrm{cos}37}{5.1}=\frac{17\times 4}{5.1}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{68}{5.1}\Rightarrow a=13.33\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

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