Question

directions: solve the problem shown below. any data, calculations, or models you use to support your thinking should be explained. problem: a vertical spring is attached to the floor, as shown in figure 1. an object of mass ( m_1 ) is placed on top of the spring and slowly lowered, compressing the spring a distance ( x_1 ) from its unstretched length, until it reaches equilibrium, as shown in figure 2. next, an object of mass ( m_2 ) is placed on top of the first object and slowly lowered, compressing the spring an additional distance ( x_2 ). the system is again in equilibrium in figure 3. what is the relationship between ( x_2 ) and ( x_1 )?

Explanation:

Step1: Apply Hooke's Law for Figure 2

In equilibrium, the spring force equals the weight of \( m_1 \). Hooke's Law is \( F = kx \), and weight is \( F_g = m_1g \). So, \( kx_1 = m_1g \). Solving for \( x_1 \), we get \( x_1=\frac{m_1g}{k} \).

Step2: Apply Hooke's Law for Figure 3

Now, the total mass is \( m_1 + m_2 \), so the spring force equals the total weight. Using Hooke's Law, \( k(x_1 + x_2)=(m_1 + m_2)g \). Expand this: \( kx_1 + kx_2 = m_1g + m_2g \).

Step3: Substitute \( kx_1 \) from Step1

From Step1, \( kx_1 = m_1g \). Substitute into the equation from Step2: \( m_1g + kx_2 = m_1g + m_2g \). Subtract \( m_1g \) from both sides: \( kx_2 = m_2g \), so \( x_2=\frac{m_2g}{k} \).

Step4: Compare \( x_2 \) and \( x_1 \)

From Step1, \( x_1=\frac{m_1g}{k} \), and from Step3, \( x_2=\frac{m_2g}{k} \). If we assume \( m_2 = m_1 \) (though not necessary, the relationship is linear in mass), but generally, the ratio \( \frac{x_2}{x_1}=\frac{m_2}{m_1} \). However, a key point: when we add \( m_2 \), the additional compression \( x_2 \) is related to \( x_1 \) by the ratio of masses. But more simply, from the force balance, for the first case, \( kx_1 = m_1g \); for the second, the total compression \( x_1 + x_2 \) gives \( k(x_1 + x_2)=(m_1 + m_2)g \). Subtracting the first equation from the second: \( kx_2 = m_2g \). So, \( x_2=\frac{m_2}{m_1}x_1 \) (since \( x_1=\frac{m_1g}{k} \), so \( \frac{m_2g}{k}=\frac{m_2}{m_1}x_1 \)). If \( m_2 = m_1 \), then \( x_2 = x_1 \), but generally, \( x_2 \) is proportional to \( m_2 \) and \( x_1 \) is proportional to \( m_1 \). However, the problem likely expects recognizing that the additional compression \( x_2 \) is related to \( x_1 \) by the ratio of masses, but a common case (if \( m_2 = m_1 \)) would be \( x_2 = x_1 \), but more accurately, from the force balance:

From \( kx_1 = m_1g \) and \( kx_2 = m_2g \), so \( \frac{x_2}{x_1}=\frac{m_2}{m_1} \). But if we consider the general relationship, the key is that the compression is proportional to the mass. So, the relationship is \( x_2=\frac{m_2}{m_1}x_1 \), or if \( m_2 = m_1 \), \( x_2 = x_1 \). But typically, in such problems, if the masses are equal, \( x_2 = x_1 \), but the derivation shows \( x_2 \) is proportional to \( m_2 \) and \( x_1 \) to \( m_1 \).

Answer:

The relationship is \( \boldsymbol{x_2 = \frac{m_2}{m_1}x_1} \) (or if \( m_2 = m_1 \), \( x_2 = x_1 \)). In general, the compression of the spring is proportional to the mass applied, so \( x_2 \) and \( x_1 \) are related by the ratio of the masses \( m_2 \) and \( m_1 \).