question_answer

A force of 750 N is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle 30° with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the frictional force acting on the block is                                   [SCRA 1994]

A)                         750 N            

B)                           500 N            

C)                         345 N             

D)                         250 N            

Correct Answer: D

Solution :

                                       Net force along the plane             = \[P-mg\sin \theta \]= \[750-500\]     = \[250\ N\]             Limiting friction = \[{{F}_{l}}={{\mu }_{s}}R={{\mu }_{s}}mg\cos \theta \]             \[=0.4\times 102\times 9.8\times cos30\text{ }=\text{ }346N\]                         As net external force is less than limiting friction therefore friction on the body will be 250 N.