A 4.52 kg block is pushed along the ceiling with a constant applied force of 85.9 N that acts at an angle of 60.5 degrees with the horizontal. the block accelerates to the right at 5.85 m/s^2. Find the coefficient of kinetic friction between the block anf the ceiling.
2007-11-11 05:59:50 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Force of friction equal to the product of the coefficient of friction u and normal to the surface force (or the force with which the object is pressed against the ceiling)
f= uN
f=u(Fsin(60.5) - mg))
The total force acting on the block is the difference between the applied force in the horizontal direction Fh and force of friction f.
Ft=Fh-f
Ft=Fcos (60.5) - u(Fsin(60.5) - mg))
Also Ft=ma
we have
u=[Fcos(60.5) - ma]/(Fsin(60.5) - mg))
2007-11-13 03:10:37 · answer #1 · answered by Edward 7 · 0⤊ 0⤋