Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Trending News

Promoted
PHYS - ENGR
Lv 4
PHYS - ENGR asked in Science & MathematicsPhysics · 10 years ago

Question about Spring constants?

A 79.0 kg student is standing atop a spring in an elevator as it accelerates upward at 3.80 m/s2. The spring constant is 2100 N/m. By how much (in m) is the spring compressed, assuming that the local acceleration due to gravity is 9.80 m/s2?

1 Answer

Relevance
  • kuiperbelt2003
    Lv 7
    10 years ago
    Favorite Answer

    the spring will record the normal force of the student standing atop it

    we find the normal force by applying newton's second law to the student

    sum of forces = m a

    normal force - weight = m a

    N - mg = ma

    N = m(g+a) = 79kg(9.8m/s/s+3,8m/s/s) = 1074.4N

    this is the force on the spring, so its compression is

    F = k x => x = F/k = 1074.4N/2100N/m = 0.51m

Still have questions? Get your answers by asking now.