- #1
Lone Wolf
- 10
- 1
- Homework Statement
- A block of mass m = 2.0 kg is on an incline plane, attached to a spring with elastic constant K = 20 N/m. The coefficient of kinetic friction between the block and the plane is equal to 1/6. Initially, the block is at rest and the spring is in the equilibrium position. Find the maximum elongation of the spring.
- Relevant Equations
- The work done by a non-conservative force is equal to the variation of mechanical energy of a system.
W(FNC) = ΔME
Work: W = F d cos(Θ)
Friction force equation: f = μN ; where μ is the coefficient of kinetic friction.
Let v be the speed of the block and x elongation of the spring beyond the equilibrium point. Initially, v = 0 and x = 0. At the maximum elongation, the block also has v = 0, it has moved a distance equal to x (parallel to the plane) and the variation of height is equal to -x⋅sin(53°).
W(FNC) = ΔME = (Kf - Ki) + (Uf - Ui)
Where K is the kinetic energy and U is the potential energy. Kf - Ki = 0 since both the initial and final speeds are zero. The potential energy is of two kinds, the gravitational potential energy and elastic potential energy. The initial elastic potential energy is zero because the spring is in the equilibrium position.
f⋅x⋅cos(180°) = m⋅g⋅Δh + 1/2 kx²
-f⋅x = m⋅g⋅(-x⋅sin(53°)) + 1/2 kx²
f = μN → f = μ⋅m⋅g⋅cos(53°))
Rearranging the terms of the equation:
0 = 2⋅m⋅g⋅x (μ⋅cos(53°) - sin(53°)) + kx²
Replacing the numerical values:
0 = 2⋅2⋅9.8⋅x⋅(1/6⋅cos(53°) - sin(53°)) + 20x²
x = 1.37 m
The person who corrected my attempt had crossed the term 1/2 kx² in the equation (whilst maintaining the remaining ones) as if it was incorrect for me to place it there. Since I am unable to get in contact with this individual to question their reasoning for doing this, I came here to see if any of you guys can help me understand why I should not have included the final elastic potential energy. It makes sense in my mind since the block is connected to the spring.