Answer to Question #140543 in Classical Mechanics for sridhar

Question

A block of mass m=2Kg is intially at rest on a horizontal surface.A horizontal force bar F_(1)=6i N and a vertical force bar
F_(2)=10j N are then applied to the block.The coefficients of static friction and kinetic friction for the block and the surface are 0.4 and 0.25 respectively.The magnitude of the frictional force acting on the block is
Ans 2.5N

Answer 1

1) Calculate the normal force:

$$N=mg-F_j=2\cdot10-10=10\text{ N}.$$

2) Find the force of static friction:

$$f_s=\mu_s N=0.4\cdot10=4\text{ N}.$$

This is the force we need to apply to make the block move. As we see, 6 N is greater than 4 N and the block will move under the horizontal force of 6 N.

3) Since the block moves, find the force of kinetic friction:

$$f_k=\mu_kN=0.25\cdot10=2.5\text{ N}.$$