A 4.0 kg 4.0kg4, point, 0, start text, k, g, end text box slides with an initial speed of 3.0 m s 3.0 s m ​ 3, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction towards a spring on a frictionless horizontal surface. When the box hits the spring, the spring compresses and stops the box momentarily. The spring constant k = 125 N m k=125 m N ​ k, equals, 125, start fraction, start text, N, end text, divided by, start text, m, end text, end fraction.

Respuesta :

Answer:

 x = 0.537 m

Explanation:

In this exercise we are asked to find the distance that the spring is compressed, let's use the concepts of conservation of energy

       

starting point. Before touching the spring

         Em₀ = K = ½ m v²

final point. When the spring has maximum compression

         [tex]E_{f}[/tex] = [tex]K_{e}[/tex] = ½ k x²

as there is no friction the mechanical energy is conserved

          Em₀ = Em_f

          ½ m v² = ½ k x²

          x = [tex]\sqrt{\frac{m}{k} }[/tex]  v

let's calculate

          x = RA[tex]\sqrt{\frac{4.0}{125} }[/tex]  3.0

          x = 0.17888 3.0

          x = 0.537 m

Answer:

0.54

Explanation:

You round the original number with two significant digits

0.537 has to be 0.54