Problem 75 - University Physics with Modern Physics

Question

CPA small block with a mass of 0.0900 \(\mathrm{kg}\) is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.75). The block is originally revolving at a distance of 0.40 \(\mathrm{m}\) from the hole with a speed of 0.70 \(\mathrm{m} / \mathrm{s} .\) The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 \(\mathrm{m} .\) At this new distance, the speed of the block is observed to be 2.80 \(\mathrm{m} / \mathrm{s}\) . (a) What is the tension in the cord in the original situation when the block has speed \(v=0.70 \mathrm{m} / \mathrm{s} ?\) (b) What is the tension in the cord in the final situation when the block has speed \(v=2.80 \mathrm{m} / \mathrm{s} ?\) (c) How much work was done by the person who pulled on the cord?

Step-by-Step Solution

Answer

(a) 0.1102 N; (b) 7.056 N; (c) 0.331 J.

Step 1: Understanding the Given Problem

We have a small block attached to a cord, moving in a circular path. Initially, it has a radius of 0.40 m and a speed of 0.70 m/s. The radius is shortened to 0.10 m, and the speed increases to 2.80 m/s. We need to calculate the tension in the cord at the initial and final states, as well as the work done by pulling the cord.

Step 2: Calculate Initial Tension

The centripetal force for circular motion is given by the formula \( F = \frac{mv^2}{r} \). For the initial situation:- Mass \( m = 0.0900 \ \mathrm{kg} \)- Initial speed \( v = 0.70 \ \mathrm{m/s} \)- Initial radius \( r = 0.40 \ \mathrm{m} \) The tension in the cord, which provides the centripetal force, is:\[ T_\text{initial} = \frac{0.0900 \times (0.70)^2}{0.40} = \frac{0.0900 \times 0.49}{0.40} \approx 0.1102 \ \mathrm{N} \]

Step 3: Calculate Final Tension

The centripetal force for the final situation is also given by \( F = \frac{mv^2}{r} \). For the final situation:- Mass \( m = 0.0900 \ \mathrm{kg} \)- Final speed \( v = 2.80 \ \mathrm{m/s} \)- Final radius \( r = 0.10 \ \mathrm{m} \)The tension in the cord is:\[ T_\text{final} = \frac{0.0900 \times (2.80)^2}{0.10} = \frac{0.0900 \times 7.84}{0.10} = 7.056 \ \mathrm{N} \]

Step 4: Determine Work Done by Pulling the Cord

Work done is equivalent to the change in kinetic energy of the block. The initial kinetic energy is:\[ KE_\text{initial} = \frac{1}{2} \times 0.0900 \times (0.70)^2 = 0.02205 \ \mathrm{J} \]The final kinetic energy is:\[ KE_\text{final} = \frac{1}{2} \times 0.0900 \times (2.80)^2 = 0.3528 \ \mathrm{J} \]The work done by pulling the cord is the difference in kinetic energies:\[ W = KE_\text{final} - KE_\text{initial} = 0.3528 - 0.02205 = 0.33075 \ \mathrm{J} \]

Key Concepts

Circular Motion

When an object moves in a path with a constant radius, it is said to be in circular motion. In such cases, the object experiences a force directed towards the center of the circle, known as centripetal force. This force keeps the object in its circular path without drifting outward. In our exercise, the block revolves around a point, maintaining a circular motion. The centripetal force is crucial because it provides the necessary force to keep the block moving in a circle.

• Initially, the block moves at a radius of 0.40 m with a speed of 0.70 m/s, maintained by the centripetal force provided by the tension in the cord.
• When the radius is reduced to 0.10 m, the speed increases to 2.80 m/s, showing that centripetal force is related to both the speed and radius.

The formula for centripetal force is given by: \[ F = \frac{mv^2}{r} \] This allows us to calculate the tension in the cord, which acts as the centripetal force keeping the block moving in a circular path.

Work and Energy

The concept of work and energy is essential to understanding how much effort is needed to change the conditions of a moving object. In the given exercise, we can measure how much work is done by observing the change in kinetic energy when altering the block's speed and path radius.

• Work is defined as the energy transfer when a force acts over a distance. Here, it occurs when we pull the cord, causing the block to speed up.
• The initial kinetic energy of the block, when it moves at 0.70 m/s, is relatively low compared to its kinetic energy at 2.80 m/s. This change signifies the amount of work done.

We can find the work done as the difference in kinetic energies: \[ W = KE_\text{final} - KE_\text{initial} \] where kinetic energy \( KE \) is given by \( \frac{1}{2} mv^2 \). The performed work reflects how pulling the cord has increased the block's energy, making it move faster.

Tension in a Cord

Tension in a cord is a force exerted through a string, rope, or in this case, a cord that allows it to pull or maintain an object in place. For an object moving in a circle, like our block, this tension provides the centripetal force needed to hold the object in its circular path.

• The initial tension supports the block at a radius of 0.40 m and speed of 0.70 m/s, calculated to be around 0.1102 N.
• When the radius is shortened to 0.10 m, and the speed increases to 2.80 m/s, the tension also increases significantly to about 7.056 N. This demonstrates how tension changes based on speed and radius.

By understanding how tension relates to these parameters, we see why it is important in maintaining circular motion and how altering the path affects the necessary tension in the cord.