Question

A 3.6 kg block is pushed along a horizontal floor by a force ModifyingAbove Upper F With right-arrow of magnitude 28 N at a downward angle θ = 40°. The coefficient of kinetic friction between the block and the floor is 0.26. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.

Answer 1

(a)

Equilibrium equation along y-axis,

-26 sin 40 - mg + N = 0

                                        N = 28 sin 40 + 3.6 * 9.8

                                           = 53.3

The frictional force is,

            f= μ N

              =0.26 *53.3

             =13.85 N

(b)

Equilibrium equation along x-axis,

                       28 cos 40 - f = ma

                  28cos 40 - f = 3.6 a

         28 cos 40 - 13.85 = 3.6 a

                                       a=2.11 m/s^2