Answer:
[tex]y_{max} = 0.829\,m[/tex]
Explanation:
Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:
[tex]v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}[/tex]
[tex]v = 2.913\,\frac{m}{s}[/tex]
The maximum compression of the spring is calculated by using the Principle of Energy Conservation:
[tex](3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}[/tex]
After some algebraic handling, a second-order polynomial is formed:
[tex]12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s[/tex]
[tex]1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0[/tex]
The roots of the polynomial are, respectively:
[tex]\Delta s_{1} \approx 0.128\,m[/tex]
[tex]\Delta s_{2} \approx -0.099\,m[/tex]
The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:
[tex]\Delta s \approx 0.128\,m[/tex]
The maximum height that the block reaches after rebound is:
[tex](3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}[/tex]
[tex]y_{max} = 0.829\,m[/tex]
Answer:
0.81 m
Explanation:
In all moment, the total energy is constant:
Energy of sistem = kinetics energy + potencial energy = CONSTANT
So, it doesn't matter what happens when the block hit the spring, what matters are the (1) and (2) states:
(1): metal block to 0.8 m above the floor
(2): metal block above the floor, with zero velocity ( how high, is the X)
Then:
[tex]E_{kb1} + E_{gb1} = E_{kb2} + E_{gs2}[/tex]
[tex]E_{kb1} + E_{gb1} = 0 + E_{gs2}[/tex]
[tex]\frac{1}{2}*m*V_{b1} ^{2} + m*g*H_{b1} = m*g*H_{b2}[/tex]
[tex]H_{b2} = \frac{V_{b1} ^{2} }{2g} + H_{b1}[/tex]
Replacing data:
[tex]H_{b2} = \frac{0.44^{2} }{2*9.81} + 0.8[/tex]
[tex]H_{b2} = \frac{0.44^{2} }{2*9.81} + 0.4[/tex]
HB2 ≈ 0.81 m
