a. F*cos20-Fk = m*a = m*0 = 0.
F*cos20-u(Fn+F*sin20) = 0
300*cos20 - u(1000+300*sin20) = 0
281.9 - 1102.6u = 0
1102.6u = 281.9
u = 0.256 = Coefficient of kinetic friction.
b. m*g = 1000N.
m = 1000/g = 1000/9.8 = 102 kg.
300*cos20 - 0.256(1000-300*sin20)=m*a
281.9 - 229.7 = m*a
m*a = 52.2
a = 52.2/m = 52.2/102 = 0.512 m/s^2.
A 1000N crate is being pushed across a level floor at a constant speed by a force F of 300N at an angle of 20.0 degrees. (a) What is the coefficient of kinetic friction between the crate and the floor? (b) If the 300-N force is instead pulling the block at an angle of 20.0 degrees above the horizontal, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).
2 answers
Thanks man you are the best
2 answers