HELP! finding distance block traveled that is attached to a spring

Question

A block of mass m = 2.5 kg is attached to a spring with spring constant k = 530 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 27° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk = 0.14. In the initial position, where the spring is compressed by a distance of d = 0.11 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero.

(search up question for diagram of problem can't include images)

If the spring pushes the block up the incline, what distance, L in meters, will the block travel before coming to rest? The spring remains attached to both the block and the fixed wall throughout its motion.

Answer 1

Answer 2

This is a conservation of energy problem: E₀(initial energy) = E_f(final energy

PE_{s =} PE_{g(gravitational potential)} + E_{lost due to friction}

Variables

F_f =µ_k F_{N =} µ_k mg cosθ

m = 2.5 kg

k = 530 N/m

θ = 27⁰

µ_k = 0.14

x(distance spring is compressed) = 0.11m

d( distance block travels up incline) = ?

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E₀(initial energy) = E_f(final energy

1. PE_{s(spring potential) =} PE_{g(gravitational potential)} + E_{lost due to friction(or heat)(or work done by friction)}
2. 1/2 kx² = mgh + (F_{f x} d) ----->µ_k mg cosθd must solve for d but h is also unknown.

to find h use sin θ = opp/hyp(h/d) ------> h =(sin 27) d since the incline makes a right triangle and d is the hypotenuse and h is the opposite side of the angle.

1. plug in numbers for all variables in the above energy conservation equation now to solve for d
2. 1/2kx² = mgh + µ_k mg cosθd
3. 1/2(530)(0.11)² = (2.5)(9.8)(sin 27 d) +(0.14)(2.5)(9.8)(cos27) d

3.2065 = 11.123d + 1.222d-------> 3.2065 = 12.345 d-----> d = 0.26 m