physics need help. Compressed a spring. figome 6 A 2kg block compressed a spring of constant 7000N/m at the bottom of a 37 degreeincline with the coefficient of kinetic friction is 0.3. If the block remains attached to the spring,how high does the block go? (If the the spring gives all its energy to the block by the time theyseparate at the spring/s equilibrium position, the block will go 1.25m above the incline )

Consider the following diagram to understand the problem: Here, the spring’s Mean position is shown with the dotted line on the incline. The spring constant and the inclination angle are denoted here by k and θ, respectively. Also, d and d’ are the block’s (mass: m) displacement in case it stays attached to the spring and in case it leaves the spring as mentioned, respectively. The vertical displacements of the block in the two cases are also shown. The block’s weight (W) is resolved along its normal reaction (N).Let m, g, and μ denote the block’s mass, the gravitational acceleration, and the given friction coefficient, respectively. The block faces the friction (say f) if it moves on the incline. The block does not move perpendicular to the incline, and hence, the total force (F) in that direction is zero. Apply Newton’s second law for the perpendicular equilibrium by taking upward force positive and determine N and f as follows: 0=∑F=N-WcosθN=mgcosθf=μmgcosθ ∵f=μNThe spring’s kinetic energy (K0) at the bottom is zero as it starts from rest. It has an elastic potential energy (say U0). If released, the block has no elastic potential energy beyond the mean position and has to do work (φ’) against the friction (f). At the top, when the block momentarily comes to rest before coming back down the incline, its kinetic energy (K’) is zero. It has a gravitational potential energy (say P’). Hence, determine the initial amplitude (A) of the spring’s harmonic motion by using the work-energy considerations as follows: K0+U0=K'+P'+φ'0+12kA2=0+mgd'sinθ+fd'12kA2=mgd'sinθ+μmgd'cosθA=2mgd'ksinθ+μcosθ=22 kg9.8 m/s21.25 m7000 N/msin37°+0.3cos37°=0.0767 mIf attached to the spring, the block has an elastic potential energy (U) at the top and has to do work (φ) against the friction (f). At the top, when the block momentarily comes to rest before coming back down the incline, its kinetic energy (K) is zero. It has a gravitational potential energy (say P). Hence, determine the displacement (d) of the spring by using the work-energy considerations as follows: K0+U0=U+K+P+φ0+12kA2=12kA-d2+0+mgdsinθ+fd12kA2=12kA2+d2-2Ad+mgdsinθ+μmgdcosθd=2kA-mgsinθ+μcosθk=27000 N/m0.0767 m-2 kg9.8 m/s2sin37°+0.3cos37°7000 N/m=0.1488 m

Answered: A 2kg block compressed a spring of…