Net Work & Work-Energy Theorem: Videos & Practice Problems

Understanding net work involves calculating the total work done by multiple forces acting on an object. The net work is the sum of individual works, calculated using the equation ${\frac{F}{d}}^{\cos }\left(\theta \right)$. The work-energy theorem relates net work to the change in kinetic energy, expressed as $W=\Delta K=K\_\mathrm{final}-K\_\mathrm{initial}$. This relationship allows for the calculation of work even when forces are unknown, emphasizing the connection between work and energy transfer.

Calculating Net Work

In physics, when calculating the net work done on an object by multiple forces, it is essential to understand that net work is the total of all individual works performed by each force acting on the object. The net work can be determined through two primary approaches, depending on the information provided in the problem.

The first method involves calculating the work done by each force using the formula:

W = F \cdot d \cdot \cos(\theta)

where W is the work done, F is the force applied, d is the displacement, and \theta is the angle between the force and the direction of displacement. After calculating the work for each force, the net work is simply the sum of these individual works:

W_{net} = W_1 + W_2 + W_3 + ... + W_n

For example, if an applied force of 15 N acts over a displacement of 10 m in the same direction, the work done is:

W_{applied} = 15 \, \text{N} \cdot 10 \, \text{m} = 150 \, \text{J}

If there is a friction force of 7 N acting in the opposite direction, the work done by friction would be:

W_{friction} = -7 \, \text{N} \cdot 10 \, \text{m} = -70 \, \text{J}

For forces acting perpendicular to the direction of motion, such as the normal force and gravitational force, the work done is zero:

W_{normal} = 0 \, \text{J}, \quad W_{gravity} = 0 \, \text{J}

Thus, the net work in this scenario would be:

W_{net} = 150 \, \text{J} + (-70 \, \text{J}) = 80 \, \text{J}

The second method involves first calculating the net force acting on the object. This is done by summing all the forces in the direction of motion. For instance, if the net force is calculated as:

F_{net} = F_{applied} + F_{friction} = 15 \, \text{N} + (-7 \, \text{N}) = 8 \, \text{N}

Once the net force is determined, the net work can be calculated using the same work formula:

W_{net} = F_{net} \cdot d \cdot \cos(\theta)

In this case, with a net force of 8 N and a displacement of 10 m, the calculation yields:

W_{net} = 8 \, \text{N} \cdot 10 \, \text{m} \cdot \cos(0) = 80 \, \text{J}

Both methods will yield the same result, demonstrating the consistency of the principles of work and force in physics. When faced with a problem, you can choose either path based on your preference or the specific requirements of the question. Understanding these concepts allows for a comprehensive approach to solving work-related problems in physics.

You pull a 3kg box on a flat surface. The coefficient of kinetic friction is 0.6. When you pull the box horizontally through a distance of 10m, it accelerates at 2m/s². Find the net work on the box.

• A. 17.6 J
• B. 35.3 J
• C. 176.4 J
• D. 60 J

Answer: (D)

To pull a 51 kg crate across a rough floor, a worker applies a force of 100 N, directed 37°above the horizontal. The coefficient of friction is 0.16. If the crate moves 3.0 m, what is the total work done on the crate?

• A. 29 J
• B. 0 J
• C. 240 J
• D. 169.6 J

Answer: (A)

The Work-Energy Theorem

Understanding the concept of work is essential in physics, as it represents the transfer of energy between objects. Work is quantified as the energy transferred when a force acts on an object, causing it to move. For instance, if you push a book and it gains 10 joules of kinetic energy, you have done 10 joules of work on that book. This leads us to the work-energy theorem, which establishes a mathematical relationship between work and 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 can be expressed with the equation:

W_net = K_final - K_initial

Where K_final is the final kinetic energy and K_initial is the initial kinetic energy. The change in kinetic energy is calculated as the difference between the final and initial kinetic energies, which can be determined using the formula:

K = \frac{1}{2}mv^2

In this context, 'm' represents the mass of the object, and 'v' represents its velocity. To illustrate this, consider a scenario where a 4-kilogram box moves from an initial speed of 6 m/s to a final speed of 10 m/s. To find the work done on the box, we can calculate the change in kinetic energy between these two points.

First, we calculate the initial kinetic energy (K_initial) at 6 m/s:

K_initial = \frac{1}{2} \times 4 \, \text{kg} \times (6 \, \text{m/s})^2 = 72 \, \text{Joules}

Next, we calculate the final kinetic energy (K_final) at 10 m/s:

K_final = \frac{1}{2} \times 4 \, \text{kg} \times (10 \, \text{m/s})^2 = 200 \, \text{Joules}

Now, we can find the net work done (W_net) on the box:

W_net = K_final - K_initial = 200 \, \text{J} - 72 \, \text{J} = 128 \, \text{Joules}

This example demonstrates the power of the work-energy theorem, allowing us to determine the total work done on an object without needing to know the specific forces involved. All that is required are the mass of the object and its speeds at two different points, highlighting the theorem's utility in solving problems in mechanics.

A box slides across the floor with an initial speed of 3.5m/s. If the coefficient of kinetic friction is 0.15, how far will the box slide before stopping completely?

• A. 1.19 m
• B. 4.17 m
• C. 9 m
• D. Not Enough Information

Answer: (B)

Net Work & Constant Speed

The Work Energy Theorem is a fundamental concept in physics that relates the work done on an object to its change in kinetic energy. In scenarios where an object moves at a constant speed, such as a 5-kilogram box being pushed on a rough surface, understanding this theorem can simplify problem-solving significantly.

In this example, the box is pushed with an applied force while experiencing kinetic friction. Since the box moves at a constant speed of 8 meters per second, it is crucial to recognize that the acceleration is zero. According to Newton's second law, the net force acting on the box can be expressed as:

$$ F_{\text{net}} = m \cdot a $$

Given that the acceleration (a) is zero, the net force (F_net) must also be zero. This indicates that the applied force is balanced by the force of friction, resulting in no net force acting on the box.

To calculate the net work done on the box, we can use the equation:

$$ W_{\text{net}} = F_{\text{net}} \cdot d \cdot \cos(\theta) $$

Substituting the net force of zero into this equation yields:

$$ W_{\text{net}} = 0 \cdot d \cdot \cos(\theta) = 0 \text{ joules} $$

This outcome illustrates that if the net work done on an object is zero, then the change in kinetic energy is also zero, as stated by the Work Energy Theorem. The kinetic energy (KE) of an object is defined as:

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

Since the speed (v) remains constant, the kinetic energy does not change, confirming that the net work done is zero. This principle is particularly useful in problem-solving; if you encounter a situation where an object is moving at a constant speed, you can immediately conclude that the net work done on the object is zero, and thus, there is no change in kinetic energy.

In summary, the Work Energy Theorem provides a powerful tool for analyzing motion. Recognizing that constant speed implies zero net work and no change in kinetic energy can streamline your approach to various physics problems.

Here’s what students ask on this topic:

What is the net work done on an object?

The net work done on an object is the total work resulting from all the forces acting on it. It is calculated by summing the individual works done by each force. Mathematically, it can be expressed as:

$W=\sum _{}^{i}W\_i$

where $W\_i$ is the work done by the $i$-th force. Each work can be calculated using the formula:

$W=Fdcos\theta$

where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors.

How do you calculate net work using the work-energy theorem?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This can be expressed as:

$W=\Delta K=K\_f-K\_i$

where $K\_f$ is the final kinetic energy and $K\_i$ is the initial kinetic energy. Kinetic energy is given by:

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

where $m$ is the mass and $v$ is the velocity. By knowing the initial and final velocities, you can calculate the change in kinetic energy and thus the net work done.

What is the relationship between work and energy transfer?

Work is a measure of energy transfer between objects. When work is done on an object, energy is transferred to or from the object. For example, if you push a book and it gains speed, the work you do on the book results in an increase in its kinetic energy. The relationship between work and energy is given by the work-energy theorem:

$W=\Delta K$

This equation shows that the net work done on an object is equal to the change in its kinetic energy, emphasizing the direct connection between work and energy transfer.

How do you calculate work done by multiple forces?

To calculate the work done by multiple forces, you need to determine the work done by each individual force and then sum them up. The work done by a single force is given by:

$W=Fdcos\theta$

where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. After calculating the work for each force, sum them up to get the net work:

$W=\sum _{}^{i}W\_i$

What is the work-energy theorem?

The work-energy theorem is a fundamental principle in physics that relates the net work done on an object to its change in kinetic energy. It states that:

$W=\Delta K=K\_f-K\_i$

where $K\_f$ is the final kinetic energy and $K\_i$ is the initial kinetic energy. This theorem highlights the direct relationship between the work done on an object and the resulting change in its kinetic energy, providing a powerful tool for solving problems involving energy transfer.