What is an ideal diode? a. The ideal diode acts as an open circuit for forward currents and as a short circuit with reverse volt
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Answer:
Approximately , assuming that:
- the height of
refers to the distance between the clay and the top of the uncompressed spring.
- air resistance on the clay sphere is negligible,
- the gravitational field strength is
, and
- the clay sphere did not deform.
Explanation:
Notations:
- Let
denote the spring constant of the spring.
- Let
denote the mass of the clay sphere.
- Let
denote the initial speed of the spring.
- Let
denote the gravitational field strength.
- Let
denote the initial vertical distance between the clay and the top of the uncompressed spring.
Let denote the maximum compression of the spring- the only unknown quantity in this question.
After being compressed by a displacement of , the elastic potential energy
in this spring would be:
.
The initial kinetic energy of the clay sphere was:
.
When the spring is at the maximum compression:
- The clay sphere would be right on top of the spring.
- The top of the spring would be below the original position (when the spring was uncompressed) by
.
- The initial position of the clay sphere, however, is above the original position of the top of the spring by
.
Thus, the initial position of the clay sphere ( above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by
.
The gravitational potential energy involved would be:
.
No mechanical energy would be lost under the assumptions listed above. Thus:
.
.
Rearrange this equation to obtain a quadratic equation about the only unknown, :
.
Substitute in ,
,
,
, and
. Let the unit of
be meters.
(Rounded. The unit of both sides of this equation is joules.)
Solve using the quadratic formula given that :
.
(The other root is negative and is thus invalid.)
Hence, the maximum compression of this spring would be approximately .