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Coefficient of Kinetic Friction... given Spring constant?
Calculate the coefficient of kinetic friction between a 4.82 kg block and the horizontal surface on which it rests if an 82.3 N/m spring must be stretched by 2.21 cm to pull it with constant speed. Assume that the spring pulls in a direction 13.4° above the horizontal.
I have tried this problem 3 different ways, assuming that the Fk=Fs and then adding in the force in they y direction as (1-sin(theta).... I am lost
2 Answers
- odu83Lv 78 years agoFavorite Answer
The force of the spring
Fs=82.3*0.0221 N
Normal force
N=m*g-Fs*sin13.4
and the horizontal dynamic
Fs*cos13.4-N*u=0 (no acceleration)
u=Fs*cos13.4/N
or
u=82.3*0.0221*cos13.4/(4.82*9.81-82.3*0.0221*sin13.4)
u=0.038
- gereckeLv 45 years ago
Please convert 2.20 Cm to 0.022 m. Spring stress = 88.2*0.022 N = a million.ninety 4 N. The spring stress works alongside the horizontal axis = a million.ninety 4 cos 14.6 N = a million.88 N. typical stress the block agains floor = 4.24*10 N = 40 2.4N. (anticipate gravity = 10 m/s²) Coefficient of kinetic friction = a million.88/40 2.4 = 0.40 4
