[" In the given diagram the "],[" frictional force acting on "6kg],[" block is "],[qquad mu=0.2k(sqrt(1)rarr10N)/(3)],[(-6)/(mu=0)]

In the above diagram calculate frictional force acting on 6 kg block

A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force vecF of magnitude 6.0 N and a vertical force vecP are then applied to the block (Fig. 6-31). The coefficients of friction for the block and surface are mu = 0.40 and mu = 0.25 . Determine the magnitude of the frictional force acting on the block if the magnitude of vecP is (a) 8.0 N. (b) 10 N, and (c) 12 N

The force acting on the block is given by F = 5 - 2t . The frictional force acting on the block after time t = 2 seconds will be: ( mu = 0.2)

The force acting on the block of mass 1 kg is given by F=5-2t . The frictional force acting on the block after time t=2 seconds will be: (mu=0.2)

A block of mass 20kg is pushed with a horizontal force of 90N . It the coefficient of static and kinetic friction are 0.4 and 0.3, the frictional force acting on the block is (g =10ms^(-2)) .

A block of mass 20kg is pushed with a horizontal force of 90N. If the coefficient of static & kinetic friction are 0.4 & 0.3, the frictional force acting on the block is: (g =10ms^(-2) )

A block of mass 20kg is pushed with a horizontal force of 90N. If the coefficient of static & kinetic friction are 0.4 & 0.3, the frictional force acting on the block is: (g =10ms^(-2) )