Work And Energy Calculating Work Done And Final Velocity In A Physics Problem

This article delves into a fascinating physics problem involving a bodybuilder pulling a 60 kg tire. We will explore the concepts of work and energy, specifically calculating the work done by the bodybuilder when pulling the tire. This scenario presents a practical application of physics principles, allowing us to understand how forces, displacement, and angles interact. The problem provides specific values for the force applied, the angle of elevation, and the distance covered, which allows for a quantitative analysis of the work done. Understanding these calculations is crucial in various fields, from sports science to engineering, where assessing the efficiency of movements and machines is paramount. By breaking down the problem step-by-step, we can clearly see how the formula for work done is applied and the significance of each variable involved. This exploration will enhance your understanding of mechanics and the role of work in transferring energy.

A bodybuilder pulls a 60 kg tire with a force of 250 N at an elevation of 20º to the horizontal.

1. If he is able to pull for 25 meters, calculate the work done by the bodybuilder.
2. Provided that the tire was initially at rest and there is frictional force of 150 N, calculate the final velocity

In this section, we will meticulously calculate the work done by the bodybuilder while pulling the tire. Work in physics is defined as the energy transferred to or from an object by the application of force along a displacement. The formula to calculate work done (${W}$) is given by:

${ W = F \cdot d \cdot cos(\theta) }$

Where:

• ${ F }$ is the magnitude of the force applied,
• ${ d }$ is the magnitude of the displacement, and
• ${ \theta }$ is the angle between the force vector and the displacement vector.

Understanding the Components

Before plugging in the values, let’s understand the components in our scenario. The bodybuilder applies a force (${ F }$) of 250 N. The tire is pulled for a distance (${ d }$) of 25 meters. The force is applied at an angle (${ \theta }$) of 20º to the horizontal. This angle is crucial because only the component of the force along the direction of motion contributes to the work done. The cosine of this angle gives us the fraction of the force that is effectively pulling the tire forward. Neglecting this angle would lead to an overestimation of the work done, as we would be considering the entire force rather than just its horizontal component. Therefore, the angle is a critical factor in accurately determining the work done in this scenario. The mass of the tire (60 kg) is not directly used in this part of the calculation, as we are focusing on the work done by the applied force over the given distance.

Applying the Formula

Now, let’s substitute the given values into the formula:

${ W = 250 \text{ N} \cdot 25 \text{ m} \cdot cos(20º) }$

To find the cosine of 20º, we can use a calculator. The cosine of 20º is approximately 0.9397. Thus, the equation becomes:

${ W = 250 \text{ N} \cdot 25 \text{ m} \cdot 0.9397 }$

Now, we multiply these values together:

${ W = 250 \cdot 25 \cdot 0.9397 \approx 5873.125 \text{ J} }$

Therefore, the work done by the bodybuilder is approximately 5873.125 Joules. This value represents the amount of energy the bodybuilder has transferred to the tire by pulling it. The unit of work is Joules (J), which is a standard unit of energy in the International System of Units (SI). This calculation highlights the direct relationship between the force applied, the distance over which it is applied, and the angle at which the force is exerted, all of which contribute to the total work done.

Conclusion

In conclusion, the work done by the bodybuilder in pulling the tire 25 meters with a force of 250 N at an angle of 20º to the horizontal is approximately 5873.125 Joules. This calculation demonstrates how to apply the work formula in a practical scenario, emphasizing the importance of considering the angle between the force and displacement vectors.

In this section, we will calculate the final velocity of the tire, considering it was initially at rest and there is a frictional force acting against the motion. To find the final velocity, we will use the work-energy theorem, which relates the work done on an object to its change in kinetic energy. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This theorem provides a powerful link between work and motion, allowing us to determine how the work done affects an object's velocity. In this case, we need to account for both the work done by the bodybuilder and the work done by the frictional force.

Understanding the Forces at Play

The bodybuilder applies a force of 250 N at an angle of 20º, as stated in the problem. However, there is also a frictional force of 150 N acting in the opposite direction of the motion. This frictional force does negative work, as it opposes the displacement of the tire. The net force acting on the tire is the difference between the horizontal component of the applied force and the frictional force. Understanding these forces is crucial because they directly influence the net work done, which in turn determines the change in kinetic energy and the final velocity of the tire. The mass of the tire (60 kg) is essential in this part of the calculation, as it is a factor in determining the kinetic energy and the effect of the net force on the acceleration of the tire.

Calculating the Net Work Done

First, we need to calculate the horizontal component of the force applied by the bodybuilder. This is given by:

${ F_x = F \cdot cos(\theta) = 250 \text{ N} \cdot cos(20º) \approx 250 \text{ N} \cdot 0.9397 \approx 234.925 \text{ N} }$

Next, we find the net force (${ F_{net} }$) by subtracting the frictional force from the horizontal component of the applied force:

${ F_{net} = F_x - F_{friction} = 234.925 \text{ N} - 150 \text{ N} = 84.925 \text{ N} }$

Now, we calculate the net work done (${ W_{net} }$) using the net force and the distance (25 meters):

${ W_{net} = F_{net} \cdot d = 84.925 \text{ N} \cdot 25 \text{ m} = 2123.125 \text{ J} }$

This net work represents the total energy transferred to the tire, taking into account the energy lost due to friction.

Applying the Work-Energy Theorem

The work-energy theorem states:

${ W_{net} = \Delta KE = KE_f - KE_i }$

Where:

• ${ \Delta KE }$ is the change in kinetic energy,
• ${ KE_f }$ is the final kinetic energy, and
• ${ KE_i }$ is the initial kinetic energy.

The kinetic energy (${ KE }$) of an object is given by the formula:

${ KE = \frac{1}{2} m v^2 }$

Where:

• ${ m }$ is the mass of the object, and
• ${ v }$ is its velocity.

Since the tire was initially at rest, its initial kinetic energy (${ KE_i }$) is zero. Thus, the equation simplifies to:

${ W_{net} = KE_f = \frac{1}{2} m v_f^2 }$

Where ${ v_f }$ is the final velocity of the tire.

Solving for the Final Velocity

We can now solve for the final velocity (${ v_f }$):

${ 2123.125 \text{ J} = \frac{1}{2} \cdot 60 \text{ kg} \cdot v_f^2 }$

Multiply both sides by 2:

${ 4246.25 = 60 \cdot v_f^2 }$

Divide both sides by 60:

${ v_f^2 = \frac{4246.25}{60} \approx 70.77 }$

Take the square root of both sides:

${ v_f = \sqrt{70.77} \approx 8.41 \text{ m/s} }$

Therefore, the final velocity of the tire is approximately 8.41 m/s.

Conclusion

The final velocity of the tire after being pulled for 25 meters, considering the frictional force, is approximately 8.41 m/s. This calculation demonstrates the application of the work-energy theorem in a scenario involving multiple forces. It highlights how the net work done, which is the difference between the work done by the applied force and the work done by friction, determines the change in kinetic energy and, consequently, the final velocity of the object. This analysis provides a comprehensive understanding of the interplay between forces, work, and energy in a real-world context.

In summary, this problem provided a detailed exploration of work and energy principles in physics. We calculated the work done by a bodybuilder pulling a tire, considering the angle of force application. Additionally, we determined the final velocity of the tire by accounting for the frictional force and applying the work-energy theorem. These calculations not only reinforce the theoretical concepts but also illustrate their practical applications. Understanding these principles is crucial for analyzing various physical scenarios, from simple mechanical systems to complex engineering designs. The problem highlights the importance of considering all forces acting on an object and their effects on its motion and energy.