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 ,
  1.  The force P is slowly increased . what is the value  of P   just  before  the block  breaks  off the floor.
  2. 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:


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

P' sin 370=5.1×10P'=53×51=5×17P'=85 N

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

ma=P cos θa=85×cos 375.1=17×45.1a=685.1a=13.33 ms

Hope this information will clear your doubts about the topic.

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