A box of mass 22g is attached to a horizontal spring with spring constant 145N/m. The spring is then extended by 6 cm, and the box is subsequently released. Calculate the speed of the box, assuming the entire elastic potential energy is converted to the box's kinetic energy.
(Give your answer to 2 decimal places if needed).

Respuesta :

Answer: 4.87 m/s

Explanation:

According to the law of conservation of energy, the total amount of energy in a system cannot be created nor destroyed. This means that the box's total elastic potential energy (U) is equivalent to its total kinetic energy (K).

Therefore, U = K.

The formula for elastic potential energy is given as:

U = [tex]\frac{1}{2}kx^{2}[/tex], where:

  • k is the spring constant (N/m)
  • x is the change in position or how much the spring is stretched (m)

Let's calculate the elastic potential energy when k = 145 and x = 0.06 (6 cm = 0.06 m):

U = [tex]\frac{1}{2}kx^{2}[/tex]

U = [tex]\frac{1}{2}(145)(0.06)^{2}[/tex]

U = 0.261J

Since U = K, 0.261 = K.

The formula for kinetic energy is given as:

K = [tex]\frac{1}{2}mv^{2}[/tex], where:

  • m is the mass of the object (kg)
  • v is the velocity of the object (m/s)

We can replace this formula for K in our equation 0.261 = K:

0.261 = [tex]\frac{1}{2}mv^{2}[/tex]

Since m is 0.022 kg (22 g = 0.22 kg):

0.261 = [tex]\frac{1}{2}(0.022)v^{2}[/tex]

Solving for v:

[tex]\frac{2(0.261)}{0.022}[/tex] = v²

v = [tex]\sqrt {\frac{2(0.261)}{0.022}}[/tex]

v = 4.87

The velocity of the box is 4.87 m/s when the box's entire elastic potential energy is converted to its entire kinetic energy.

Learn more about the law of conservation of energy here: https://brainly.com/question/41690512

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