Asked by Rachel
You push a block of mass m against a horizontal spring, compressing the spring a distance x. Then you release the block, and the spring sends it sliding across a tabletop. It stops a distance d from where you release it. Let k be the spring constant. What is the coefficient of kinetic friction between the block and the table? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
μk =
μk =
Answers
Answered by
crystal
Use conservation of energy E = K + U
Ui (initial potential from spring) = Kf (final kinetic energy
1/2 kx^2 = 1/2 mv^2
v = x*sqrt(k/m)
From kinematics:
d = (Vf^2-Vi^2) / (2a)
d = (x*sqrt(k/m))^2 / (2a) (substituting our v from above)
d = (kx^2) / (2*ma)
We know from forces that friction = µ*N and N = mg
the net force on the object after leaving the spring will be that from friction, therefore F = µmg = ma
µmg = (kx^2) / (2d)
µ = (kx^2) / (2dmg)
Ui (initial potential from spring) = Kf (final kinetic energy
1/2 kx^2 = 1/2 mv^2
v = x*sqrt(k/m)
From kinematics:
d = (Vf^2-Vi^2) / (2a)
d = (x*sqrt(k/m))^2 / (2a) (substituting our v from above)
d = (kx^2) / (2*ma)
We know from forces that friction = µ*N and N = mg
the net force on the object after leaving the spring will be that from friction, therefore F = µmg = ma
µmg = (kx^2) / (2d)
µ = (kx^2) / (2dmg)
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.