Physics, asked by nusrathcassim, 2 months ago

a person holds a 4.0 kg block at position a shown above on the left, in contact with an uncompressed vertical spring with a spring constant of 500 n/m. ther person gently lowers the block from rest at position a to rest at position b. which of the following describes the change in the energy of the block-spring-earth system as a result of the block being lowered?​

Attachments:

Answers

Answered by adil589
1

Answer:

The force of each name is 1 kilogram, a fraction of a fraction on an equal block. In this state of Mukila, is the value index of the three blocks across the fourth block and the outpost contact block with the

Answered by aidanlee808
4

Answer:

Decrease by ~1.5 J

Explanation:

Position A, the spring is uncompressed so it does not have potential spring energy in the moment, only potential gravitational energy or "mgh". mgh for position A = 11.76. Position B is compressed by 0.1 m from equilibrium (0.3m). So it has mgh+(1/2kx^2). mgh=7.84 and 1/2kx^2 = 250 x 0.1^2 = 2.5.

2.5 + 7.84 = 10.34

11.76 - 10.34 = 1.42 which is closest to 1.5 J (note it is lost energy because height decreases)

Similar questions