The spring-mounted 0.84-kg collar A oscillates along the horizontal rod, which is rotating at the constant angular rate rad/s. At a certain instant, r is increasing at the rate of 610 mm/s.
If the coefficient of kinetic friction between the collar and the rod is 0.61, calculate the friction force F exerted by the rod on the collar at this instant.

Friction = 5.027 N

Explanation:

Kinetic friction does not change with speed or acceleration. The only thing that is required is to state that the object is moving.

Solving the force equation for friction we get:

Friction = coefficient of friction* normal force exerted by the object

Friction = u * R

here R is = 0.84 * Gravity = 0.84 * 9.81 = 8.2404 N

u = 0.61 (as stated in the question)

So frictional force will be:

Friction = 0.61 * 8.2404

Friction = 5.027 N

i already answered it by 34/s

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr