Asked by Alex
A sled is pulled with a horizontal force of 17 N along a level trail, and the acceleration is found to be 0.40 m/s2. An extra mass m = 4.4 kg is placed on the sled. If the same force is just barely able to keep the sled moving, what is the coefficient of kinetic friction between the sled and the trail?
Answers
Answered by
drwls
Let the friction force be f without the added mass. It increases to
f (M+4.4)/M with the added mass.
You have two unknowns (f and M) and need two equations.
17 - f = M*a = M*0.40
17 - f*(M+4.4)/M = (M+4.4)*0 = 0 (zero acceleration with added mass)
f*(M+4.4)/M -Ff = 0.4 M
f*4.4/M = 0.4M
f = M^2/11
17-f = 17 - M^2/11 = 0.4M
Solve for M and use that to solve for f. Take the postitive root when solving for M.
The coefficient of kinetic friction is
muk = f/(M*g)
f (M+4.4)/M with the added mass.
You have two unknowns (f and M) and need two equations.
17 - f = M*a = M*0.40
17 - f*(M+4.4)/M = (M+4.4)*0 = 0 (zero acceleration with added mass)
f*(M+4.4)/M -Ff = 0.4 M
f*4.4/M = 0.4M
f = M^2/11
17-f = 17 - M^2/11 = 0.4M
Solve for M and use that to solve for f. Take the postitive root when solving for M.
The coefficient of kinetic friction is
muk = f/(M*g)
Answered by
Alex
That makes a lot more sense now. Thanks!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.