Problem 63 - University Physics with Modern Physics

Question

A 3.00 -kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of \(T(t)=(36.0 \mathrm{N} / \mathrm{s}) t .\) The box is at rest at \(t=0 .\) The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at ( i \(t=1.00 \mathrm{s}\) and (ii) \(t=3.00 \mathrm{s} ?\) (b) What is the maximum distance that the box descends below its initial position? (c) At what value of \(t\) does the box return to its initial position?

Step-by-Step Solution

Answer

(a) (i) -3.81 m/s at 1.00s, (ii) 36.09 m/s at 3.00s. (b) Max descent: 4.02 m. (c) Return at 2.45s.

Step 1: Understand the Forces

The box is subject to two forces: the tension in the rope, given by \(T(t) = (36.0 \, \text{N/s}) \, t\), and the gravitational force, which is constant and equal to \(mg\) where \(m = 3.00 \, \text{kg}\) and \(g = 9.81 \, \text{m/s}^2\). This means the gravitational force is \(29.43 \, \text{N}\).

Step 2: Set Up the Net Force Equation

The net force on the box is given by \(F_{net} = T(t) - mg\). This becomes \(F_{net} = (36.0t - 29.43) \, \text{N}\).

Step 3: Use Newton’s Second Law to Find Acceleration

According to Newton’s second law, \(F = ma\), so \((36.0t - 29.43) = 3.00a\). Solving for \(a\), the acceleration of the box is \(a = 12.0t - 9.81\).

Step 4: Integrate Acceleration to Find Velocity

To find the velocity, integrate acceleration over time: \(v(t) = \int (12.0t - 9.81) \, dt\). This yields \(v(t) = 6.0t^2 - 9.81t + C\), where \(C\) is the integration constant. Since \(v(0) = 0\), \(C = 0\). Thus, \(v(t) = 6.0t^2 - 9.81t\).

Step 5: Calculate Velocity at Specific Times

Substitute \(t = 1.00\) s into the velocity equation: \(v(1.00) = 6.0(1.00)^2 - 9.81(1.00) = -3.81 \, \text{m/s}\). Substitute \(t = 3.00\) s: \(v(3.00) = 6.0(3.00)^2 - 9.81(3.00) = 36.09 \, \text{m/s}\).

Step 6: Integrate Velocity to Find Position

The position \(x(t)\) can be found by integrating the velocity function: \(x(t) = \int (6.0t^2 - 9.81t) \, dt = 2.0t^3 - 4.905t^2 + C\). Since \(x(0) = 0\), \(C = 0\). So, \(x(t) = 2.0t^3 - 4.905t^2\).

Step 7: Find Maximum Descend Below Initial Position

The maximum descent occurs when the velocity is zero (change in direction). Solve \(v(t) = 6.0t^2 - 9.81t = 0\): Factoring gives \(t(6.0t - 9.81) = 0\). Ignoring \(t = 0\), solve \(6.0t = 9.81\) to find \(t = 1.635 \, \text{s}\). Substitute into position equation: \(x(1.635) = 2.0(1.635)^3 - 4.905(1.635)^2 \approx -4.02 \, \text{m}\).

Step 8: Find Return Time to Initial Position

Set the position equation equal to zero and solve for \(t\): \(2.0t^3 - 4.905t^2 = 0\) which factors to \(t^2(2.0t - 4.905) = 0\). Disregarding \(t = 0\), solve \(2.0t = 4.905\) to find \(t = 2.4525 \, \text{s}\).

Key Concepts

Tension in Rope

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.

Net Force

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.

Integration of Acceleration

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

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.