Thursday
March 23, 2017

Post a New Question

Posted by on Thursday, September 29, 2011 at 9:40pm.

A student places her 442 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 4.44 cm, then releases the book. What is the book's speed as it slides away? The spring constant is 1102 N/m.

  • Physics - , Wednesday, April 18, 2012 at 8:20pm

    (1/2)*K*x^2 = (1/2)*m*v^2
    K-Spring constant, 1102 N/m
    x-Compression distance, 4.44cm = .0444m
    m-Mass of physics book, 442 g = .442 Kg

    Re-arrange:

    v = x*(K/m)^0.5
    v = .0444*(1102/.442)^0.5
    v = 2.22 m/s

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question