Static mu = 198N/(55 kg*9.8m/s^2)
Kinetic mu = 175/(55*9.8)
Must be pretty rough ice. Get out the Zambomi.
A 55 kg ice skater is at rest on a flat skating rink. A 198 N horizontal forceis needed to set the skater in motion.However, after the skater is inmotion, a horizontal force of 175 N keeps the skater moving at a constantvelocity. Find the coefficients of static and kinetic friction between theskates and the ice.
5 answers
static = 0.37
kinetic = 0.32
kinetic = 0.32
no
Given :
M= 55kg Fsmax= 198N.
Kinetic friction= 175
Unknown:
Coefficients of kinetic friction = ?
Coefficient of static friction = ?
Coefficient of kinetic friction:
Kinetic friction (Fk) / Normal friction (Fn)
(Fn=mg)
Therefore: 175 / (55)(9.81) = 0.32
Coefficient of static friction:
Fsmax / Normal friction (Fn)
Therefore: 198 / (55)(9.81) = 0.37
Good luck on exams🤝
M= 55kg Fsmax= 198N.
Kinetic friction= 175
Unknown:
Coefficients of kinetic friction = ?
Coefficient of static friction = ?
Coefficient of kinetic friction:
Kinetic friction (Fk) / Normal friction (Fn)
(Fn=mg)
Therefore: 175 / (55)(9.81) = 0.32
Coefficient of static friction:
Fsmax / Normal friction (Fn)
Therefore: 198 / (55)(9.81) = 0.37
Good luck on exams🤝
Thank you!
1 answer