Question

A block of mass 5 kg is placed on a horizontal surface with coefficient of friction 0.2, then the maximum and minimum value of force F for which the block remains at rest.

Answer 1

Explanation:

F minimum =f

F =umg

F=0.2 ×5×10

F=10N

so it os option 1

Answer 2

Answer:

15 N . 10 N.

Explanation:

The minimum amount of force is given as fmin = μmg where μ is the coefficient of friction and the value of minimum force is given so because the force which is just require ti overcome the weight of the block itself and the friction so that it moves is given as μmg = 0.2*5*10 which is 10 N.

Since, the force of 15N is given and the minimum force is not greater than the applied force hence the maximum force is 15 N.