Problem 71 - University Physics with Modern Physics

Question

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (\(\textbf{Fig. P6.71}\)). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed \(\upsilon = 0.70\) m/s? (b) What is the tension in the cord in the final situation, when the block has speed \(\upsilon = 2.80\) m/s? (c) How much work was done by the person who pulled on the cord?

Step-by-Step Solution

Answer

(a) 0.0735 N (b) 4.704 N (c) 0.2205 J

Step 1: Understanding the Initial Tension Calculation

To find the tension in the cord when the block is revolving at a speed of 0.70 m/s and radius 0.40 m, we use the centripetal force formula. The tension force provides the necessary centripetal force, so it equals \( T = \frac{m v^2}{r} \). Here, \( m = 0.0600 \text{ kg} \), \( v = 0.70 \text{ m/s} \), and \( r = 0.40 \text{ m} \).

Step 2: Calculating the Initial Tension

Substitute the known values into the centripetal force formula: \[T = \frac{0.0600 \times (0.70)^2}{0.40}\]Calculate to find:\[T = \frac{0.0600 \times 0.49}{0.40} = \frac{0.0294}{0.40} = 0.0735 \text{ N}\]So, the tension in the cord initially is 0.0735 N.

Step 3: Understanding the Final Tension Calculation

For the final scenario, we again use the formula for centripetal force with the new speed and radius of the revolution. The tension in the cord when \( v = 2.80 \text{ m/s} \) and \( r = 0.10 \text{ m} \) is given by\[T = \frac{m v^2}{r}.\]

Step 4: Calculating the Final Tension

Substitute the new values into the centripetal force formula:\[T = \frac{0.0600 \times (2.80)^2}{0.10}\]Calculate to find:\[T = \frac{0.0600 \times 7.84}{0.10} = \frac{0.4704}{0.10} = 4.704 \text{ N}\]Thus, the tension in the cord in the final situation is 4.704 N.

Step 5: Understanding Work Done by the Force

Work done by the person pulling the cord can be found using the work-energy principle, which states that the work done is the change in kinetic energy. Calculate the initial and final kinetic energies and find their difference. Initial kinetic energy:\[ K_i = \frac{1}{2} m v_i^2 \]where \( v_i = 0.70 \text{ m/s} \).Final kinetic energy:\[ K_f = \frac{1}{2} m v_f^2 \]where \( v_f = 2.80 \text{ m/s} \).

Step 6: Calculating Initial and Final Kinetic Energies

Calculate the initial kinetic energy:\[K_i = \frac{1}{2} \times 0.0600 \times (0.70)^2 = \frac{1}{2} \times 0.0600 \times 0.49 = 0.0147 \text{ J}\]Calculate the final kinetic energy:\[K_f = \frac{1}{2} \times 0.0600 \times (2.80)^2 = \frac{1}{2} \times 0.0600 \times 7.84 = 0.2352 \text{ J}\]

Step 7: Calculating the Work Done

The work done (W) by pulling the cord is the change in kinetic energy:\[W = K_f - K_i = 0.2352 \text{ J} - 0.0147 \text{ J} = 0.2205 \text{ J}\]So, the work done by the person is 0.2205 J.

Key Concepts

Kinetic Energy

Kinetic energy is an essential concept in understanding how energy moves with objects as they speed up or slow down. It reflects the energy an object possesses due to its motion. For a given object, the kinetic energy is calculated using the formula:\[ K = \frac{1}{2} mv^2 \]where:- *m* is the mass of the object (in kilograms),- *v* is the velocity of the object (in meters per second).In our scenario, a small block moves at varying speeds depending on how far it is from the hole it's revolving around. Initially, the block has a kinetic energy of 0.0147 J when revolving at 0.70 m/s.When the radius decreases and its speed increases to 2.80 m/s, the block's kinetic energy spikes to 0.2352 J. This showcases how significantly kinetic energy enhances as speed multiplies, demonstrating that kinetic energy is very sensitive to changes in speed due to the squared factor of velocity.

Tension

Tension is a force exerted by a string or rope when it is pulled tight by forces acting from opposite ends. In circular motion, such as that of a block attached to a cord, tension provides the necessary centripetal force for the block to maintain its circular path.For each scenario, tension is computed as:\[ T = \frac{mv^2}{r} \]where:- *m* is the mass,- *v* is the velocity,- *r* is the radius of rotation.In the initial setting, tension is calculated to be 0.0735 N at a speed of 0.70 m/s with a radius of 0.40 m. However, when the radius is reduced to 0.10 m and speed increases to 2.80 m/s, tension drastically rises to 4.704 N. This shows how tension, directly related to speed and inversely related to radius, increases sharply with a decrease in radius and increase in velocity.

Work-Energy Principle

The work-energy principle bridges the concept of work done on an object and its kinetic energy change. It states that the work done by forces on an object results in a change in its kinetic energy. The work done can be calculated by:\[ W = \Delta K = K_f - K_i \]where:- \( K_f \) is the final kinetic energy,- \( K_i \) is the initial kinetic energy.In the exercise involving the rotating block, the work done by pulling the cord from below alters its kinetic energy from 0.0147 J to 0.2352 J. As a result, the work done is measured as 0.2205 J. This perfectly illustrates how an increase in kinetic energy, due to increasing speed, reflects the work applied on the block, thereby making the block move faster along a tighter circle.