Answer:
(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.
Explanation:
Initially the spring is at equilibrium,
Work done by all forces = change in kinetic energy
Work = ∇K.E
Work = Kf -Ki =0
Since the work done = 0 since the body is at rest.
W(spring) + W(gravity) = 0
W(spring) + W(gravity) = 0
W(spring) = -W(gravity)
Work done by the block on the spring = W(block/spring)
W(block/spring) + W(spring) = 0
W(spring) = -∫kx.dx
W(spring) = ½k(X²i - X²f) ; Xi =0, Xf = 15.3cm = 0.153m
W(spring) = -½* 460 * (0.153)²
W(spring) = - 5.38NM
Work done by block on spring = + 5.38NM
(b). Workdone by spring on the block = -5.38NM.
Note: This is so because the displacement of the force is in the opposite direction to the previous one since they counter each other to maintain equilibrium.
(C). W(spring) +W(gravity) = 0
½kx² + mg(h + x) = 0
-5.83 + mg(h + 0.153) =0
5.83 = 0.425*9.8 (h + 0.153)
5.83 = 4.165(h + 0.153)
H = 1.399 - 0.153
H = 1.246m
(D).
If the release height was 6ho
H = 6* 1.246m = 7.476m
W(spring) = W(gravity)
½kx² = mg(7.476 + x)
Note: At maximum compression, the blocks would be at rest.
½Kx² = mg(h + x)
½ * 460 * x² = 0.425 * 9.8 * (7.476 + x)
230x² = 4.165 (7.476 + x)
230x² = 31.137 + 4.165x
230x² - 4.165x - 31.137 = 0
Solving the quadratic equation ( i would suggest you use formula method for easy navigation of the variables)
X = + 0.3771m or -0.3589m
But we can't have a negative compression value,
X = + 0.3771m
