Calculate the amount of potential energy that was initially stored in the spring. g= 9.8?

Question

A wooden block with mass 1.85 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 34.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.95 m up the incline from A, the block is moving up the incline at a speed of 7.30 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is mu_k = 0.550. The mass of the spring is negligible.

Answer 1

Total energyTE(A) = PE(A) = the potential energy of the spring before release (point A).

Total energy TE(B) = PE(B) + KE(B) + WE(B); potential, kinetic, and work energies at point B. PE(B) = mgh = mgS sin(theta); where m = 1.85 kg, g = 9.8 m/sec^2, S = 6.95 m, and theta = 34 deg. KE(B) = 1/2 mv^2; where v = 7.3 mps. And WE(B) = kmg cos(theta) S; where k = .55 the coefficient of friction.

From the conservation of energy TE(A) = PE(A) = mgS sin(theta) + 1/2 mv^2 + kmg cos(theta) S = PE(B) + KE(B) + WE(B). Thus, PE(A) = mgS sin(theta) + 1/2 mv^2 + kmg cos(theta) S and all the factors on the RHS of the equation are given; so you can do the math.

Here's the physics. By the conservation of energy law, the total potential energy of the compressed spring has to be the same as the total potential, kinetic, and friction energy of the block just as it leaves the decompressed spring.