91Ó°ÊÓ

When dealing with springs, the spring constant is a crucial concept defined by Hooke's Law. Hooke's Law tells us that the force exerted by a spring is directly proportional to the displacement it is subjected to. Mathematically, this is expressed as \( F = kx \), where \( F \) is the force applied, \( x \) is the displacement from the equilibrium position, and \( k \) is the spring constant.

• The spring constant, \( k \), is a measure of the spring's stiffness. A higher \( k \) value indicates a stiffer spring.
• Units for the spring constant are Newtons per meter (N/m).

In our exercise, a 25 N force compresses the spring by 18 cm, equated to 0.18 meters. By rearranging Hooke's Law, \( k = \frac{F}{x} \), we calculated \( k = \frac{25}{0.18} = 138.89 \, \text{N/m} \). This value helps in determining how the spring will behave when released.

Friction plays a significant role when objects move across surfaces. It acts in the opposite direction of motion, essentially resisting it. The frictional force is quantified using the equation \( f = \mu mg \), where \( \mu \) is the coefficient of friction, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity.

• The frictional force is dependent on both the nature of the surfaces in contact and the mass of the sliding object.
• For this exercise, the coefficient of friction, \( \mu \), is given as 0.30.

The block weighs 180 grams, which converts to 0.180 kg. With gravity being \( 9.8 \, \text{m/s}^2 \), the frictional force can be calculated as \( f = 0.30 \times 0.180 \times 9.8 = 0.5292 \, \text{N} \). This frictional force opposes the movement of the block as it extends on the table.

The principle of energy conservation is fundamental in physics, stating that energy cannot be created or destroyed, only transformed. In the context of our exercise, the energy initially stored in the compressed spring transforms into work done against friction and the energy in the stretched spring. The equation for energy conservation in this problem is given by:\[\frac{1}{2}kx_i^2 = f(d_i + d_f) + \frac{1}{2}kd_f^2\]

• On the left-hand side, \( \frac{1}{2}kx_i^2 \) represents the initial potential energy of the compressed spring.
• On the right-hand side, \( f(d_i + d_f) \) is the work done against friction, and \( \frac{1}{2}kd_f^2 \) is the final potential energy when the spring is stretched.

Initially, the spring has potential energy calculated from its compression, 2.250 J in this case (step 5). Solving the equation allows us to find the stretched distance \( d_f \), which is approximately 0.10 m. This result confirms how far the spring stretches beyond its equilibrium after accounting for energy losses due to friction.