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koban [17]
3 years ago
12

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact
Physics
2 answers:
Anton [14]3 years ago
4 0

Answer:

y_{max} = 0.829\,m

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:

v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}

v = 2.913\,\frac{m}{s}

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

(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}

After some algebraic handling, a second-order polynomial is formed:

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

1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_{1} \approx 0.128\,m

\Delta s_{2} \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

The maximum height that the block reaches after rebound is:

(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}

y_{max} = 0.829\,m

Amiraneli [1.4K]3 years ago
3 0

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)

<u>Then:</u>

<u />E_{kb1} + E_{gb1}  = E_{kb2} + E_{gs2}<u />

<u />E_{kb1} + E_{gb1}  = 0 + E_{gs2}<u />

<u />\frac{1}{2}*m*V_{b1} ^{2}   + m*g*H_{b1}  = m*g*H_{b2}<u />

<u />H_{b2}  =  \frac{V_{b1} ^{2} }{2g}  + H_{b1}<u />

<u>Replacing data:</u>

<u />H_{b2}  =  \frac{0.44^{2} }{2*9.81}  + 0.8<u />

<u />H_{b2}  =  \frac{0.44^{2} }{2*9.81}  + 0.4<u />

HB2 ≈ 0.81 m

<u />

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