91Ó°ÊÓ

When dealing with springs, understanding Hooke's Law is key. It provides the basis for calculating the force exerted by a spring when it is compressed or stretched. Hooke's Law is expressed as:

• \( F_{s} = k \cdot d \)

This formula indicates that the spring force, \( F_{s} \), is directly proportional to the displacement, \( d \), and the spring constant, \( k \).
In the given problem, the spring constant is 20 lb/ft and the spring is compressed by 0.2 ft. By substituting these values into the formula, we find that the spring exerts a force of:

• \( F_{s} = 20 \cdot 0.2 = 4 \, \text{lb} \)

This force acts against the friction and is critical in moving the blocks once they are released.

Friction plays a significant role when an object moves across a surface. Kinetic friction is the force that opposes this movement. It is calculated using the equation:

• \( F_{f} = \mu_{k} \cdot N \)

Here, \( F_{f} \) is the force of friction, \( \mu_{k} \) is the coefficient of kinetic friction, and \( N \) is the normal force.
For both blocks in this exercise, the normal force equals the weight, since there is no vertical movement. This makes the calculation straightforward:
For block A:

• \( F_{fA} = 0.2 \cdot 8 = 1.6 \, \text{lb} \)

For block B:

• \( F_{fB} = 0.2 \cdot 6 = 1.2 \, \text{lb} \)

These values help us understand how much force is needed to overcome friction and initiate the movement of the blocks.

Understanding Newton’s Second Law is essential for calculating acceleration. It links the concepts of force, mass, and acceleration through the equation:

• \( F_{net} = m \cdot a \)

Where \( F_{net} \) is the net force acting on the object, \( m \) is the mass, and \( a \) is the acceleration.
For block A, the net force is calculated by subtracting the frictional force from the spring force:

• \( F_{net}^A = 4 \, \text{lb} - 1.6 \, \text{lb} = 2.4 \, \text{lb} \)

The mass of block A can be derived from its weight, taking the weight divided by gravitational acceleration (\( 32.2 \, \text{ft/s}^2 \), where 1 lb equals \( 1 \, \text{lb}/{32.2 \, \text{ft/s}^2} \)). Thus, the acceleration \( a_A \) is calculated as:

• \( a_A = \frac{2.4}{8/32.2} \approx 9.34 \, \text{ft/s}^2 \)

Similarly, for block B:

• \( F_{net}^B = 4 \, \text{lb} - 1.2 \, \text{lb} = 2.8 \, \text{lb} \)
• \( a_B = \frac{2.8}{6/32.2} \approx 14.01 \, \text{ft/s}^2 \)

These calculations provide the acceleration values that define how quickly each block will start moving once released.