Answer:
v = 0.84m/s, v(max)= 0.997m/s
Explanation:
Initial work done by the spring, where c is the compression = 0.28m:
[tex]W_s = \frac{1}{2}kc^2[/tex]
Work lost to friction:
[tex]W_f =\mu mg(c-x)[/tex]
Energy:
[tex]E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2[/tex]
(a) Solve for v:
[tex]v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}[/tex]
(b) Solve [tex]\frac{dv}{dx}=0[/tex] for x:
[tex]\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}[/tex]
[tex]\frac{dv}{dx}=0[/tex] if:
[tex]\mu g-\frac{k}{m}x = 0[/tex]
[tex]x_{max} = \frac{\mu gm}{k}[/tex]
[tex]v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}[/tex]
Elastic potential energy is the energy stored due to the deformation of an elastic object
(a) The speed at the instant the box leaves the spring is approximately 0.84 m/s
(b) The maximum speed of the box is approximately 0.996 m/s
Reason:
Known parameters are;
Mass of the box, m = 1.60 kg
The compression of the spring, x = 0.280 m
The force constant of the spring, K = 48.0 N/m
Coefficient of kinetic friction between the box and the horizontal surface, [tex]\mu_k[/tex] = 0.300
(a) The potential energy of the spring = 0.5 × K × d²
The normal reaction of the block on the surface, N = 1.60 kg × 9.81 m/s² = 15.696 N
The friction force, [tex]F_f[/tex] = N × [tex]\mu_k[/tex]
[tex]F_f[/tex] = 15.696 N × 0.30 = 4.7088 N
Work done by friction, [tex]W_f[/tex] = [tex]F_f[/tex] × (d - x)
We get;
0.5 × K × d² - [tex]F_f[/tex] × (d - x) = 0.5·m·v² + 0.5 × K × x²
Which gives;
28 × 0.28² - 4.7088×(0.28 - x) = 0.5 × 1.6 × v² + 24·x²
1.8816 - 1.318464 + 4.7088·x = 0.8·v² + 24·x²
0.563136 + 4.7088·x = 0.8·v² + 24·x²
0.563136 + 4.7088·x - 24·x² = 0.8·v²
Therefore, when x = 0, we have;
0.563136 + 4.7088*0 - 24*0² = 0.8·v²
Which gives;
0.563136 = 0.8·v²
[tex]v = \sqrt{\dfrac{0.563136}{0.8} } \approx 0.84[/tex]
The speed at the instant the box leaves the spring, v ≈ 0.84 m/s
(b) The maximum speed is given as follows;
0.563136 + 4.7088·x - 24·x² = 0.8·v²
[tex]v =\sqrt{ 0.70392+ 5.886 \cdot x - 30 \cdot x^2}[/tex]
At the maximum velocity, [tex]v_{max}[/tex], we have;
[tex]\dfrac{dv}{dx } = 0 = \dfrac{d}{dx} \left(\sqrt{ 0.70392+ 5.886 \cdot x - 30 \cdot x^2}\right) = \dfrac{5.886- 60 \cdot x}{2 \times \sqrt{ 0.70392+ 5.886 \cdot x - 30 \cdot x^2} }[/tex]
Which gives;
5.886 - 60·x = 0
5.886 - 60·x
x = 0.0981
[tex]\therefore v_{max} =\sqrt{ 0.70392+ 5.886 \times 0.0981 - 30 \times 0.0981^2} \approx 0.996[/tex]
The maximum speed of the box during the motion, [tex]v_{max}[/tex] ≈ 0.996 m/s
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