91Ó°ÊÓ

Kinematics is the branch of physics dealing with the motion of objects without considering the forces behind the motion. In this exercise, kinematics helps us describe the motion of the ice block on a frictionless surface through its position-time equation. The position of the block as a function of time is given by the equation:

• \( x(t) = \alpha t^2 + \beta t^3 \)

To find the velocity, we differentiate the position function with respect to time:

• \( v(t) = \frac{d}{dt}(\alpha t^2 + \beta t^3) = 2\alpha t + 3\beta t^2 \)

Velocity is a necessary parameter to understand how fast the block is moving at any given moment. By substituting the time \(t = 4\,s\), you find the velocity to be \(v(4) = 1.60 \, m/s\). The value showcases the speed and direction of the block at a specific time point in its motion.

Newton's Second Law of Motion is a fundamental principle that explains how the force applied to an object is proportional to the mass and acceleration of that object. The law is expressed with the formula:

• \( F = ma \)

In this problem, the force \(\vec{F} \) applied by the worker is explored through this principle. First, you derive the acceleration expression from the velocity function:

• \( a(t) = \frac{d}{dt}(2\alpha t + 3\beta t^2) = 2\alpha + 6\beta t \)

Substituting the parameters \(t = 4\,s\), \(\alpha = 0.200\,m/s^2\), and \(\beta = 0.0200\,m/s^3\), the acceleration at that moment is \( a(4) = 0.560 \, m/s^2\). With a mass of \( m = 6.00 \, kg\), Newton's Second Law provides the force equation \( F = 6.00 \times 0.560 = 3.36 \, N\). This calculation allows us to know the magnitude of the force that the worker needs to apply to achieve this specific motion.

The Work-Energy Principle states that the work done on an object is equal to the change in its kinetic energy. The formula used is:

• \( W = \Delta KE = \frac{1}{2}mv_{final}^2 - \frac{1}{2}mv_{initial}^2 \)

Kinetic energy depends on both mass and velocity, highlighting how motion translates into energy. In the scenario of the block, its initial velocity at time \( t=0 \) is zero, simplifying our calculation:

• \( v_{initial} = 0 \)
• \( W = \frac{1}{2}(6.00)(1.60^2 - 0) \)

The work done by the force through the first 4 seconds is calculated to be \( 7.68 \, J\). This quantifies the energy transferred in moving the block over this period, bridging the gap between force application and movement of the block.