becomes: Ef = Ei - losses → mgdsin θ = ½ kX2 - fk d
The friction force fk = μk N = μk mg cos θ
The energy equation becomes: ½ kX2 = mgdsin θ + μk mg dcos θ
Solving for d gives:
| d | = |
|
What is d if kinetic friction is present?
Use D0EL.
Step 1: Define/draw diagram and coordinate system.
Step 2: Choose a consistent zero.
Step 3: Energy conservation
Step 4: Losses.
The previous energy conservation equation:
mgdsin θ = ½ kX2
The friction force fk =
μk N = μk mg cos θ
The energy equation becomes:
½ kX2 = mgdsin θ +
μk mg dcos θ
Solving for d gives:
becomes: Ef = Ei - losses
→ mgdsin θ = ½ kX2 - fk d
d
=
kX2
2mg [sin θ + μkcos θ]