Respuesta :
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
Answer:
a) Work done by the block on the spring [tex]W_{b}[/tex] = 5.38 Joules
b) Work done by the spring on the block [tex]W_{s}[/tex] = - 5.38 Joules
c) [tex]h_{0} = 1.14 m[/tex]
d) Maximum compression of the spring, x = 0.36 m
Explanation:
Spring constant, k = 460 N/m
mass, m = 0.425 kg
compression, x= 15.3 cm = 0.153 m
a) Work done by the block on the spring, [tex]W_{b}[/tex]= [tex]0.5kx^{2}[/tex]
[tex]W_{b}[/tex] = [tex]0.5*460*0.153^{2}[/tex]
[tex]W_{b}[/tex] = 5.38 Joules
b) Work done by the spring on the ball, [tex]W_{s}[/tex] = - (Work done by the block on the spring, [tex]W_{b}[/tex])
[tex]W_{s}[/tex] = - [tex]W_{b}[/tex]
[tex]W_{s}[/tex] = - 5.38 Joules
c) Value of [tex]h_{0}[/tex]
Applying the principle of energy conservation
[tex]mgh_{0} = mgx + 0.5kx^{2}[/tex]
Inserting the appropriate values into the energy conservation equation above
[tex]0.425*9.8*h_{0} = -(0.425*9.8*0.153) + (0.5*460*0.153^{2} )\\4.165h_{0} = 6.0213\\h_{0} = 4.75/4.165\\h_{0} = 1.14 m[/tex]
d)Maximum compression if height = 6h₀
[tex]mg(6h_{0}) = -mgx + 0.5kx^{2}\\0.425*9.8*(6*1.14) = -(0.425*9.8x) + (0.5*460*x^{2})\\28.49 = -4.165x + 230x^{2}\\[/tex]
Find the value of x using the quadratic formula
a = 230, b = -4.165, c = -28.49
[tex]x = \frac{-b \pm \sqrt{b^{2}-4ac } }{2a} \\x = \frac{4.165 \pm \sqrt{(-4.165)^{2}-(4*230*(-28.49) } }{2*230}\\x = \frac{4.165 \pm 161.95}{460} \\[/tex]
[tex]x = \frac{4.165 +161.95}{460}\\[/tex] = [tex]0.36[/tex]
or [tex]x = \frac{4.165 -161.95}{460}\\[/tex] [tex]= -0.34[/tex]
x = 0.36 m is the only realistic value
