91Ó°ÊÓ

Tension in a rope is the pulling force exerted by the rope on an object. In this exercise, the tension is a time-dependent force given by the function \( T(t) = (36.0 \, \text{N/s}) \, t \). This means the tension increases linearly with time. The value \( 36.0 \, \text{N/s} \) dictates how rapidly the tension changes.
To understand the role of tension, consider that it must overcome gravity for the box to accelerate upwards. If the tension is less than the gravitational force at a given time, the box will continue to descend. As time progresses, tension increases, leading to a change in the box's motion as it begins to move upwards once the tension surpasses the gravitational pull.

The net force is the overall force acting on the box, determined by the difference between the tension in the rope and gravitational force. It can be expressed by the equation \( F_{\text{net}} = T(t) - mg \). With \( mg = 29.43 \, \text{N} \), this becomes:

• \( F_{\text{net}} = (36.0t - 29.43) \, \text{N} \)

Understanding net force is pivotal because it dictates the box’s acceleration (as per Newton's Second Law). At times when \( F_{\text{net}} \) is positive, the box will accelerate upward. If negative, the box continues to move downward. This interplay between tension and gravity heavily influences the motion of the box.

To determine the velocity of an object, we first need to integrate the acceleration. From the net force, we found the acceleration \( a \) using Newton's second law: \( F = ma \) leading to \( a = 12.0t - 9.81 \). We now integrate this acceleration:

• \( v(t) = \int (12.0t - 9.81) \ dt \)
• This results in the velocity function: \( v(t) = 6.0t^2 - 9.81t + C \)

The constant \( C \) can be determined by considering the initial condition \( v(0) = 0 \), which gives \( C = 0 \). Thus, the velocity function simplifies to \( v(t) = 6.0t^2 - 9.81t \). Integration here allows for precise tracking of how fast and in which direction the box is moving at any given time.

Velocity calculation involves substituting specific time values into the velocity function to find out how fast the box is moving at that moment. For instance:

• At \( t = 1.00 \, \text{s} \), the velocity \( v(1.00) = 6.0(1.00)^2 - 9.81(1.00) = -3.81 \, \text{m/s} \)
• This negative value shows the box is still moving downwards since gravity still overpowers tension.

Similarly:

• At \( t = 3.00 \, \text{s} \), substituting gives \( v(3.00) = 6.0(3.00)^2 - 9.81(3.00) = 36.09 \, \text{m/s} \)
• This positive velocity indicates the box is moving upward, as now tension has become strong enough to overcome gravity.

The changes in velocity over time illustrate how external forces like tension can significantly influence an object’s speed and direction.