Question

Question:(II) A vertical spring (ignore its mass), whose spring constant is\({\bf{875}}\;{\bf{N/m}}\)attached to a table and is compressed down by 0.160 m. (a) What upward speed can it give to a 0.380-kg ball when released? (b) How high above its original position (spring compressed) will the ball fly?

Short answer

(a) The upward speed of the ball provided by the spring when released is 7.88 m/s.

(b) The ball can fly above its original position at the height of 3 m.

Step-by-step solution

Understanding the vertical motion of the spring

The spring is fully compressed; so the elastic potential energy is zero and hence it has only kinetic energy. When the spring is uncompressed, it has some potential energy and kinetic energy.

According to the conservation of energy, when the spring is compressed, the kinetic energy is equal to the sum of the spring potential energy, and the potential energy when the spring is uncompressed.

Identification of given data

The given data can be listed below as:

• The compression distance of the spring is\({x_2} = 0.160{\rm{ m}}\).
• The mass of the ball is\(m = 0.380{\rm{ kg}}\).
• The spring stiffness constant is\(k = 875{\bf{ }}{\rm{N/m}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Representation of the diagram of the vertical spring

The diagram can be shown as:

Here, k is the spring constant, \({x_1}\) is the uncompressed distance of the spring, and (1), (2), and (3) are the cases of the spring.

(a) Determination of the upward speed of the ball, the spring can give to the ball when released.

The initial point is considered when the spring is compressed. The final point is considered when the spring is uncompressed.

The total energy of the first case is equal to the total final energy of the second case. The energy can be expressed as:

\(\begin{array}{c}{E_1} = {E_2}\\K.{E_1} + P.{E_1} + S.{E_1} = K.{E_2} + P.{E_2} + S.{E_2}\\0 + mg{x_2} + \frac{1}{2}kx_2^2 = \frac{1}{2}m{v^2} + 0 + 0\\\frac{1}{2}m{v^2} = \frac{1}{2}kx_2^2 + mg{x_2}\end{array}\)

Here, consider case (1). The potential energy due to gravity denoted as\(P.{E_1}\)is equal to zero.\(S.{E_1}\)is the spring potential energy,\(K.{E_1}\)is the kinetic energy, which is equal to zero, and\(P.{E_1}\)is the potential energy of the spring compressed, which is equal to zero.

Here, consider case (2).\(K.{E_2}\)is the kinetic energy of the ball,\(S.{E_2}\)is the final spring potential energy, which is zero at the uncompressed (final) position, \(P.{E_2}\)is the potential energy due to gravity, and\(v\)is the velocity of the ball.

This can be further solved as:

\(\begin{array}{c}{v^2} = \frac{2}{{2m}}kx_2^2 + \frac{{2mg{x_2}}}{m}\\{v^2} = \frac{1}{m}kx_2^2 + 2g{x_2}\\v = \sqrt {\left( {\frac{1}{m}kx_2^2 + 2g{x_2}} \right)} \end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}v = \sqrt {\left( {\frac{1}{{0.380{\rm{ kg}}}} \times 875{\bf{ }}{\rm{N/m}}\left( {\frac{{1{\rm{ kg/}}{{\rm{s}}^2}}}{{1{\rm{ N/m}}}}} \right) \times {{\left( {0.160{\rm{ m}}} \right)}^2} + 2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 0.160{\rm{ m}}} \right)} \\ = \sqrt {62.086} {\rm{ m/s}}\\ = 7.88{\rm{ m/s}}\end{array}\)

Thus, the upward speed of the ball provided by the spring when released is 7.88 m/s.

(b) Determination of the distance between the initial position and final position of the spring

The spring has no elastic potential energy or kinetic energy at the uncompressed position or final position. It has gravitational potential energy and spring potential energy due to the uncompressed distance of the spring.

According to the conservation of energy, the elastic potential energy of the spring is equal to the potential energy of the ball. This can be expressed as:

\(\begin{array}{c}{E_1} = {E_3}\\K.{E_1} + P.{E_1} + S.{E_1} = K.{E_3} + P.{E_3} + S.{E_3}\\K.{E_1} + S.{E_1} = K.{E_3} + \left( {P.{E_3} - P.{E_1}} \right) + S.{E_3}\\0 + \frac{1}{2}kx_1^2 = 0 + mg{x_1} + 0\end{array}\)

This can be further solved as:

\(\begin{array}{c}\frac{1}{2}kx_2^2 = mg{x_1}\\{x_1} = \frac{{kx_2^2}}{{2mg}}\end{array}\)

Here, consider case (3).\(K.{E_3}\)is the kinetic energy of the ball when it flies in the air,\(S.{E_3}\)is the spring potential energy, which is zero at the uncompressed (final) position, and\(P.{E_3}\)is the potential energy due to gravity.

Substitute the values in the above expression.

\(\begin{array}{c}{x_1} = \frac{1}{{2 \times 0.380{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}}} \times 875{\bf{ }}{\rm{N/m}}\left( {\frac{{1{\rm{ kg/}}{{\rm{s}}^2}}}{{1{\rm{ N/m}}}}} \right) \times {\left( {0.160{\rm{ m}}} \right)^2}\\ = \frac{{22.4}}{{7.46}}{\rm{ m}}\\ = 3{\rm{ m}}\end{array}\)

Thus, the ball can fly above its original position at the height of 3 m.