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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 relaxe
Question
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
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Physics
3 years
2021-07-20T21:42:46+00:00
2021-07-20T21:42:46+00:00 2 Answers
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Answers ( )
Answer:
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:
The maximum compression of the spring is calculated by using the Principle of Energy Conservation:
After some algebraic handling, a second-order polynomial is formed:
The roots of the polynomial are, respectively:
The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:
The maximum height that the block reaches after rebound is:
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:
Replacing data:
HB2 ≈ 0.81 m