Question

A mass m is attached to a spring which is held stretched a distance x by a force F, Fig. 6–45, and then released. The spring pulls the mass to the left, towards its natural equilibrium length. Assuming there is no friction, determine the speed of the mass m when the spring returns: (a) to its normal length (x = 0); (b) to half its original extension \(\left( {{x \mathord{\left/{\vphantom {x 2}} \right.} 2}} \right)\).

FIGURE 6–45 Problem 75.

Short answer

(a)The mass moves at speed\(\sqrt {\frac{{Fx}}{m}} \)when it returns to\(x = 0\).

(b) The mass moves at speed \(\sqrt {\frac{{3Fx}}{{4m}}} \) when it returns to \(x = \frac{x}{2}\).

Step-by-step solution

Calculation of elastic potential energy

The elastic potential energy value can be determined by analyzing the spring force constant and the stretch distance of the spring.

It varies linearly to the value of the stretch distance for a constant value of force. So, as the value of the stretch distance of the spring increases, the spring's potential energy also increases.

Calculate the speed of the mass when the spring returns to its normal length

(a)

According to the work-energy theorem, the work done is equal to the change in the kinetic energy. Therefore, the expression for the work done can be written as:

\(\begin{aligned}{W_{{\rm{net}}}} &= \Delta KE\\ &= \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2\end{aligned}\)

Here,\(m\)is the mass,\({v_1} = v\)is the initial speed of the mass, and\({v_2} = 0\)is the final speed of the mass.

According to the work-energy theorem, the work done by the spring force will be equal to the change in the kinetic energy. Therefore, you can write:

\(\begin{aligned}{W_{\rm{s}}} &= \Delta KE\\ - \frac{1}{2}k{x^2} &= 0 - \frac{1}{2}m{v^2}\\k{x^2} &= m{v^2}\end{aligned}\)… (i)

Here,\(x\)is the stretched distance and\(k\)is the spring constant.

The expression for the spring force is given by:

\(\begin{aligned}F &= kx\\k &= \frac{F}{x}\end{aligned}\)… (ii)

Substitute the value of equation (ii) in equation (i).

\(\begin{aligned}k{x^2} &= m{v^2}\\\left( {\frac{F}{x}} \right){x^2} &= m{v^2}\\m{v^2} &= Fx\\v &= \sqrt {\frac{{Fx}}{m}} \end{aligned}\)

Thus, the mass moves at speed \(\sqrt {\frac{{Fx}}{m}} \) when it returns to \(x = 0\).

Calculate the speed of the mass when the spring returns to half its original extension

(b)

According to the law of conservation of energy, the total mechanical energy of the system remains conserved.

As the mass returns to the state\(x = \frac{x}{2}\), the total spring potential energy gets converted into the kinetic energy of the mass and the potential energy of the spring. Therefore,

\(\begin{aligned}{E_1} &= {E_2}\\\frac{1}{2}k{x^2} &= \frac{1}{2}m{v^2} + \frac{1}{2}k{\left( {\frac{x}{2}} \right)^2}\\m{v^2} &= k{x^2} - k{\left( {\frac{x}{2}} \right)^2}\\m{v^2} &= \frac{{4k{x^2} - k{x^2}}}{4}\\{v^2} &= \frac{{3k{x^2}}}{{4m}}\end{aligned}\)

Substitute the value of equation (ii) in the above equation.

\(\begin{aligned}{v^2} &= \frac{{3\left( {\frac{F}{x}} \right){x^2}}}{{4m}}\\v &= \sqrt {\frac{{3Fx}}{{4m}}} \end{aligned}\)

Thus, the mass moves at speed \(\sqrt {\frac{{3Fx}}{{4m}}} \) when it returns to \(x = \frac{x}{2}\).