91Ó°ÊÓ

Kinematics is the branch of physics that describes the motion of objects without considering the forces causing the motion. In the given exercise, we are interested in the equation of motion of a crate that moves under the influence of an applied force over time.
The equation provided, \( y(t) = (2.80 \, \text{m/s})t + (0.610 \, \text{m/s}^3)t^3 \), tells us about the height of the crate as a function of time, \( t \). This is a position-time relationship in one-dimensional motion.

• The term \( (2.80 \, \text{m/s})t \) is straightforward; it represents linear motion at a constant velocity.
• The term \( (0.610 \, \text{m/s}^3)t^3 \) indicates an additional velocity component that changes over time, causing the motion to be non-linear.

Understanding the components of the equation helps in analyzing how quickly and dynamically the crate’s position is changing as time progresses.

Differentiation is a vital tool in physics for understanding how quantities change over time. In this exercise, we use differentiation to analyze the crate's movement by first deriving its velocity and then its acceleration from the position-time equation.
Differentiating the position function \( y(t) \) gives the velocity \( v(t) \), helping us understand how fast the crate is moving at any point in time. The expression becomes:\[ v(t) = \frac{dy}{dt} = 2.80 + 3 \cdot 0.610 \cdot t^2 \]
This calculation not only reveals the velocity itself but also shows that the velocity is time-dependent and changes with acceleration.

Next, we differentiate the velocity to obtain acceleration \( a(t) \), describing how the velocity changes over time: \[ a(t) = \frac{dv}{dt} = 3 \cdot 0.610 \cdot 2t \] At the specific time \( t = 4.00 \text{s} \), we substitute the value to find the exact velocity and acceleration at that moment, which are crucial for further force calculations.

In the context of this problem, accelerated motion is characterized by the changing velocity of the crate due to both the constant force and the time-dependent term in the height equation. When the crate is suspended by a rope, the acceleration affects its upward motion, considering both the applied force and the gravitational pull.
Newton’s Second Law of Motion tells us that the force acting on an object is the product of its mass and acceleration \( F = m \, a \). However, the total upward force needed, \( F_t \), must overcome gravity to achieve the net acceleration:\[ F_t - mg = ma \]
Therefore, the total force is obtained by substituting the known values into the adjusted formula: \( F(t) = m(a + g) \), where \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \).

• Once the exact acceleration from earlier is known,\( a(4) = 14.64 \, \text{m/s}^2 \), it becomes straightforward to compute the total force exerted on the crate using this equation.
• This shows how forces result from and relate to accelerated motion, providing a clear idea of how various factors like velocity change and gravitational force play a role in determining net force on objects in motion.