A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0 degrees with the horizontal. The block accelerates forward at 6.00m/s/s. determine the coefficient of kinetic friction between the block and the ceiling.
2007-12-26 12:10:45 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
85*cos(55)-(85*sin(55)-4*9.81)*µk=4*6
solve for µk
0.81
j
2007-12-26 12:37:46 · answer #1 · answered by odu83 7 · 1⤊ 0⤋
two equations two unknows T1coS(ANGLE OF FIRST ) = T2 cos ( angle of second ) T1sin (angle) + T2 sin (angle of secon rope ) = 598 you didnt give any angle this time here are two eq and if you know the angles you will have two unknowns you can easily find two T1 and T2
2016-04-11 02:04:58 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Given:
Fnet = ma
Fx - Ff = ma
Fa(cos55.0deg) - uk(Fn) = ma
85.0N(cos55.0deg)-uk(85.0N*sin55.0deg)=4.00 kg(6.00m/s^2)
uk = [85.0N(cos55.0deg)-4.00kg(6.00m/s^2)]/85.0Nsin55.0deg
uk = 0.36 ANS
teddy boy
2007-12-26 12:45:05 · answer #3 · answered by teddy boy 6 · 0⤊ 1⤋