Ask Question
10 March, 21:05

A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. The upper end of the spring is depressed 0.10 m, and a 2.0-kg block is placed against the depressed spring. The system is then released from rest. What is the kinetic energy of the block at the instant it has traveled 0.10 m and the spring has returned to its uncompressed length?

+4
Answers (1)
  1. 11 March, 01:02
    0
    kinetic energy is 2.02 J

    Explanation:

    Given data

    k = 600 N/m

    angle = 30°

    spring = 0.10 m

    mass = 2 kg

    to find out

    kinetic energy of the block

    solution

    we know initial energy that is

    Energy 1 = 1/2 * kx²

    put here k = 600 and x = 0.1

    energy 1 = 1/2 * 600 (0.1) ²

    energy = 3

    potential energy PE = mgh

    PE = 2 * 9.8 * 0.1sin30

    PE = 0.98 J

    so we know final energy is

    E = KE of block + PE

    so

    kinetic energy = 3 - 0.98

    kinetic energy = 2.02 J
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A spring (k = 600 N/m) is at the bottom of a frictionless plane that makes an angle of 30° with the horizontal. The upper end of the spring ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers