Question

A 3.00-kg box that is several hundred meters above the earth's surface 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 results in a tension in the rope of $T(t)=$ (36.0 N/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 s and (ii) $t=$ 3.00 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?

Short answer

Velocity is -3.8 m/s at 1 s, 28.2 m/s at 3 s; max descends 1.65 m; returns at 2.45 s.

Step-by-step solution

Define the forces acting on the box

The two forces acting on the box are the tension $T(t)$ in the rope and the gravitational force $mg$, where $m=3.0$ kg (mass of the box) and $g=9.8$ m/s² (acceleration due to gravity). Hence, the net force $F(t)$ is given by $F(t)=T(t)-mg$.

Substitute the tension in the force equation

Given $T(t)=36.0t$ N, and substituting $m=3.0$ kg, we apply this in the net force equation: $F(t)=36.0t-3.0\times 9.8$. This results in $F(t)=36.0t-29.4$.

Apply Newton's second law

Newton's second law states that $F=ma$. Substituting our expression for $F(t)$ gives $ma=36.0t-29.4$. With $m=3.0$ kg, we find $a=\frac{36.0t-29.4}{3.0}$.

Calculate acceleration as a function of time

Evaluating the acceleration, $a(t)=12t-9.8$.

Integrate to find velocity function

To find velocity $v(t)$, integrate $a(t)=12t-9.8$: $v(t)=\int (12t-9.8)dt=6t^2-9.8t+C$. Since the box starts from rest, $v(0)=0$, so $C=0$. Therefore, $v(t)=6t^2-9.8t$.

Calculate velocity at given times

For (i) $t=1.00$ s, $v(1)=6(1{)}^2-9.8(1)=-3.8$ m/s. For (ii) $t=3.00$ s, $v(3)=6(3{)}^2-9.8(3)=28.2$ m/s.

Integrate to find the position function

To find position $y(t)$, integrate $v(t)=6t^2-9.8t$: $y(t)=\int (6t^2-9.8t)dt=2t^3-4.9t^2+C$. Initially, $y(0)=0$, so $C=0$. Thus, $y(t)=2t^3-4.9t^2$.

Find maximum descent and return time

To find maximum descent, solve $v(t)=0$: $6t^2-9.8t=0$ gives $t=0$ or $t=\frac{9.8}{6}=1.633$ s. Use $y(t)=2t^3-4.9t^2$ to find height at $t=1.633$. For return time, solve $y(t)=0$ for positive $t$ excluding 0.

Solve for the exact descent and return times

Plugging $t=1.633$ into $y(t)$, $y(1.633)\approx -1.65$ m is the maximum descent. Solve $2t^3-4.9t^2=0$: we know $t$ is $t=0$ or solve $2t^2-4.9t=0$ giving $t\approx 2.45$ s as the return time.

Key concepts

Forces on an Object

When an object is in motion, multiple forces can act upon it. Understanding how these forces interact is a crucial aspect of physics. In the case of a 3.00 kg box suspended by a rope several meters above the Earth, the key forces at play are tension and gravitational force.

Newton's Second Law applies here, stating that the net force acting on an object is equal to the product of its mass and acceleration, expressed as $F=ma$. The net force $F(t)$ affecting the box varies with time as the tension changes. Therefore, the relationship can be defined by the equation $F(t)=T(t)-mg$, where:

• $T(t)$ is the time-dependent tension in the rope.
• $mg$ is the gravitational force acting downward.

By analyzing these forces, we can determine how the box accelerates or decelerates, affecting its motion over time.

Tension in a Rope

The tension in a rope is a pulling force that occurs within the rope when it is subjected to an external force. In this scenario, the tension is not constant but changes with time, represented by the equation $T(t)=36.0t$ N.

This increasing tension is applied upward, countering the gravitational pull on the suspended box. The tension's time dependency implies that as time progresses, the force exerted by the rope increases linearly with time. This change in tension has a direct influence on the box's acceleration.

• Initially, the box is at rest, and the only significant force is gravity.
• As tension increases, it first slows the box's descent and eventually pulls it upward.

Understanding the role of tension is vital for explaining the box's motion, as it directly affects both acceleration and velocity.

Gravitational Force

Gravitational force is a fundamental force acting on all objects with mass. It is the attractive force between the object and the Earth. In our problem, the gravitational force exerted on the box is $mg$, where $m=3.0$ kg (the box's mass) and $g=9.8$ m/s² (acceleration due to Earth's gravity).

Gravitational force consistently pulls the box downward regardless of other forces. It's crucial for determining the net force because it directly opposes the tension in the rope. This interplay between gravity and tension dictates the box's movement.

• The gravitational force remains constant at $29.4$ N ($3.0\times 9.8$).
• Initially, gravity causes the box to descend as tension builds up slowly.
• Once tension surpasses gravitational force, the direction of box's motion changes.

Thus, gravitational force is essential in understanding both the descent and subsequent ascent of the box.