4. A 2-kilogram block is dropped from a height of 0.45 meter above an uncompressed spring, as shown above. The spring has a spring constant of 200 N/m and negligible mass. The block strikes the end of the spring and sticks to it. a. Determine the speed of the block at the instant it hits the end of the spring. +9=½v² 1.8=√²² (v = 1.34 16 m/s) U₂ = k mgh mu² b. Determine the force in the spring when the block reaches the equilibrium position 20=-200-X X= alm my-kx F=my (200 Z c. Determine the distance that the spring is compressed at the equilibrium position -> (.1m d. Determine the speed of the block at the equilibrium position ½kx²=12²-200 = 4 Y = 1/2mv² 8=2v² v=2m/ 0.45 m ldu 2 kg - 200 N/m e. Is the speed of the block a maximum at the equilibrium position, explain. Yes, it is the point where spring. An instant earlier and it is still accelerating, loter it is decelerating there is no force being applied by the

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Could you check my work on D and B
(D) The kinetic energy of the block is at a maximum at x = A.
(E) The kinetic energy of the block is always equal to the potential energy of the spring.
4. A 2-kilogram block is dropped from a height of 0.45 meter above an
uncompressed spring, as shown above. The spring has a spring constant of 200 N/m
and negligible mass. The block strikes the end of the spring and sticks to it.
a. Determine the speed of the block at the instant it hits the end of the spring.
U₂ = k mgh mu² *9=½/2v² 1.8= √² (v = 1.3416 m/s)
b. Determine the force in the spring when the block reaches the equilibrium position
20=-200-X X = 1m
F=my
=(200
my-kx
c. Determine the distance that the spring is compressed at the equilibrium position
•Im
d. Determine the speed of the block at the equilibrium position
Y = 1/2mv ²
½kx²=-2²-200=4
T
0.45 m
2 kg
- 200 N/m
8=2v²v-2m/
e. Is the speed of the block a maximum at the equilibrium position, explain.
it is the
Yes,
Point where there is no force being applied by the
spring. An instant earlier and it is still accelerating, loter it is decelerating.
Transcribed Image Text:(D) The kinetic energy of the block is at a maximum at x = A. (E) The kinetic energy of the block is always equal to the potential energy of the spring. 4. A 2-kilogram block is dropped from a height of 0.45 meter above an uncompressed spring, as shown above. The spring has a spring constant of 200 N/m and negligible mass. The block strikes the end of the spring and sticks to it. a. Determine the speed of the block at the instant it hits the end of the spring. U₂ = k mgh mu² *9=½/2v² 1.8= √² (v = 1.3416 m/s) b. Determine the force in the spring when the block reaches the equilibrium position 20=-200-X X = 1m F=my =(200 my-kx c. Determine the distance that the spring is compressed at the equilibrium position •Im d. Determine the speed of the block at the equilibrium position Y = 1/2mv ² ½kx²=-2²-200=4 T 0.45 m 2 kg - 200 N/m 8=2v²v-2m/ e. Is the speed of the block a maximum at the equilibrium position, explain. it is the Yes, Point where there is no force being applied by the spring. An instant earlier and it is still accelerating, loter it is decelerating.
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