Posted by megan on .
A 3.20 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0° with the horizontal.
The block accelerates to the right at 7.50 m/s2.
Determine the coefficient of kinetic friction between block and ceiling.
Would appreciate some guidance on how the Normal force works in this case
Fb = 3.2kg * 9.8 = 31.36N = ver(normal)
force of block.
Fh = 85*cos55 = 48.75N.
Fv = 31.36 + 85*sin55 = 101N = ver. force.
Fn = Fh - uFv = ma,
48.75 - 101u = 3.2 * 7.5 = 24,
48.75 -101u = 24,
-101u = 24 - 48.75,
-101u = -24.75,
u = -24.75 / -101 = 0.245 = coefficient
NOTE: Fn = Net force.