Intro to Calculating Work: Videos & Practice Problems

Work (W) is the transfer of energy, calculated using the equation W=Fd⁢cos(θ), where F is the force, d is the displacement, and θ is the angle between them. Positive work occurs when the force aids motion, while negative work occurs when it opposes motion. For gravity, work is positive when falling and negative when rising, illustrating energy transfer in kinetic energy (KE) as KE = 1/2 × mv^2.

Work Done by a Constant Force

Understanding the concept of work is essential in physics, particularly in relation to energy and forces. Work, denoted by the symbol w, is defined as the transfer of energy that occurs when a force is applied to an object, causing it to move. The fundamental relationship between work and kinetic energy can be illustrated through a simple example: when a box at rest on a frictionless surface is pushed, it begins to move, gaining velocity and thus kinetic energy. The energy imparted to the box comes from the force exerted by the person pushing it.

The unit of work is the joule (J), which is equivalent to the energy transferred when a force of one newton moves an object one meter in the direction of the force. The mathematical expression for calculating work done by a constant force is given by the equation:

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

In this equation, F represents 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. It is important to note that both force and displacement should be considered as positive values when calculating work.

For instance, if a 2-kilogram box is pulled with a force of 3 newtons over a distance of 5 meters, the work done can be calculated as follows:

\( W = 3 \, \text{N} \cdot 5 \, \text{m} \cdot \cos(0^\circ) = 3 \cdot 5 \cdot 1 = 15 \, \text{J} \)

In this case, since the force and displacement are in the same direction, the work done is positive.

Conversely, if a 5-kilogram cart moving to the right with an initial velocity of 10 meters per second experiences a stopping force of 100 newtons acting to the left, the work done by this stopping force can be calculated similarly. Here, the displacement is 2.5 meters to the left, and the angle between the force and displacement is 180 degrees:

\( W = 100 \, \text{N} \cdot 2.5 \, \text{m} \cdot \cos(180^\circ) = 100 \cdot 2.5 \cdot (-1) = -250 \, \text{J} \)

This negative value indicates that the work done by the stopping force is taking energy away from the cart, which is consistent with the idea that work can be either positive or negative. Positive work occurs when the force aids the motion of the object, while negative work occurs when the force opposes the motion.

In summary, work is a crucial concept that describes the transfer of energy through the application of force over a distance. Understanding how to calculate work and its implications on energy transfer is fundamental in the study of physics.

You pull a 5kg box vertically up with a constant 100N force for 2m. How much work do you do?

• A. 51 J
• B. 100 J
• C. 102 J
• D. 200 J
• E. 500 J
• F. 1000 J

Answer: (D)

Work by a 2D Push/Pull

In this scenario, we analyze the work done on a 10-kilogram suitcase being pulled across the floor with an applied force of 50 newtons over a distance of 20 meters. The key concept here is understanding how to calculate work, which is defined as the product of force, distance, and the cosine of the angle between the force and the direction of displacement. The formula for work is given by:

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

In this case, the angle between the applied force and the direction of movement is 37 degrees. Plugging in the values, we calculate the work done by the applied force:

W = 50 \, \text{N} \cdot 20 \, \text{m} \cdot \cos(37^\circ) = 800 \, \text{J}

This result indicates that 800 joules of work is done by the applied force in moving the suitcase.

Next, we decompose the applied force into its horizontal (F_x) and vertical (F_y) components. The horizontal component is calculated using:

F_x = F \cdot \cos(\theta) = 50 \cdot \cos(37^\circ) \approx 40 \, \text{N}

For the vertical component, we use:

F_y = F \cdot \sin(\theta) = 50 \cdot \sin(37^\circ) \approx 30 \, \text{N}

Now, we calculate the work done by the horizontal component (F_x). Since F_x acts in the same direction as the displacement, the angle between them is 0 degrees, leading to:

W_x = F_x \cdot d \cdot \cos(0^\circ) = 40 \, \text{N} \cdot 20 \, \text{m} \cdot 1 = 800 \, \text{J}

This confirms that the horizontal component is responsible for the work done on the suitcase, yielding the same result of 800 joules.

Finally, we consider the work done by the vertical component (F_y). Since F_y acts vertically while the displacement is horizontal, the angle between them is 90 degrees, resulting in:

W_y = F_y \cdot d \cdot \cos(90^\circ) = 30 \, \text{N} \cdot 20 \, \text{m} \cdot 0 = 0 \, \text{J}

This outcome is logical, as the vertical force does not contribute to the horizontal movement of the suitcase. Thus, the total work done on the suitcase is solely attributed to the horizontal component of the applied force.

Work Done By Friction

In this problem, we analyze the work done by various forces acting on a box weighing 50 newtons. It's important to note that the weight (mg) is already given as 50 newtons, so we do not need to convert this to mass in kilograms. The normal force, which acts vertically and balances the weight, is also 50 newtons since these are the only two vertical forces acting on the box.

As the box is pulled to the right, kinetic friction acts to the left, opposing the motion. The force of kinetic friction (F_k) can be calculated using the formula:

F_k = μ_k × N

where μ_k is the coefficient of kinetic friction (0.7 in this case) and N is the normal force (50 newtons). Thus, we find:

F_k = 0.7 × 50 = 35 newtons.

To calculate the work done by the kinetic friction, we use the equation:

W_k = F_k × d × cos(θ)

Here, d is the distance traveled (8 meters), and θ is the angle between the force of friction and the displacement. Since the friction force acts in the opposite direction to the displacement, θ is 180 degrees. The cosine of 180 degrees is -1, leading to:

W_k = 35 × 8 × cos(180°) = 35 × 8 × (-1) = -280 joules.

This negative work indicates that friction opposes the motion, which is a consistent characteristic of frictional forces.

Next, we calculate the work done by the weight force (mg). The work done by gravity is given by:

W_g = mg × d × cos(θ)

In this case, the weight force acts downward while the displacement is horizontal, making the angle θ equal to 90 degrees. The cosine of 90 degrees is 0, so:

W_g = 50 × 8 × cos(90°) = 0 joules.

This result is expected, as the weight force does not contribute to the work done in the horizontal direction.

Finally, we consider the work done by the normal force (N). Similar to the weight force, the normal force acts vertically upward while the displacement is horizontal, resulting in the same angle of 90 degrees. Thus, we have:

W_N = N × d × cos(θ) = 50 × 8 × cos(90°) = 0 joules.

In summary, the work done by kinetic friction is -280 joules, while the work done by both the weight force and the normal force is 0 joules. This illustrates how forces can do positive, negative, or zero work depending on their direction relative to the displacement of the object.

Work Done By Gravity

Understanding the work done by gravity is essential in physics, particularly when analyzing the motion of objects under its influence. Work is defined as the transfer of energy, and it can be either positive or negative depending on the direction of the force relative to the motion of the object. When the force aligns with the motion, the work is positive; conversely, when the force opposes the motion, the work is negative.

To calculate the work done by gravity, we use the formula:

$$ W_g = F \cdot d \cdot \cos(\theta) $$

where:

• $$ W_g $$ is the work done by gravity,
• $$ F $$ is the gravitational force (which is $$ mg $$, where $$ m $$ is mass and $$ g $$ is the acceleration due to gravity, approximately $$ 9.8 \, \text{m/s}^2 $$),
• $$ d $$ is the displacement of the object, and
• $$ \theta $$ is the angle between the force and the displacement vector.

In scenarios where an object is falling, such as a book falling from a shelf, the gravitational force acts downward, and the displacement is also downward. Here, the angle $$ \theta $$ is 0 degrees, making the cosine term equal to 1. Thus, the work done by gravity can be simplified to:

$$ W_g = mg \cdot \Delta y $$

For example, if a 5.1 kg book falls 2 meters, the work done by gravity is:

$$ W_g = 5.1 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 2 \, \text{m} = 100 \, \text{J} $$

This positive work indicates that the gravitational force assists the motion of the book as it falls.

Additionally, the kinetic energy of the book just before it hits the ground can be calculated using the formula for kinetic energy:

$$ KE = \frac{1}{2} mv^2 $$

In this case, the kinetic energy at the moment before impact is also found to be 100 joules, demonstrating the principle that work done by gravity translates into kinetic energy.

When analyzing objects moving upwards, such as a rock thrown vertically, the gravitational force still acts downward while the displacement is upward. Here, the angle between the force and displacement is 180 degrees, resulting in:

$$ W_g = -mg \cdot \Delta y $$

For instance, if a 2 kg rock is thrown upwards with an initial velocity of 15 m/s and reaches a height of 11.5 meters, the initial kinetic energy is:

$$ KE_{initial} = \frac{1}{2} \cdot 2 \, \text{kg} \cdot (15 \, \text{m/s})^2 = 225 \, \text{J} $$

Calculating the work done by gravity as the rock ascends gives:

$$ W_g = -2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 11.5 \, \text{m} = -225 \, \text{J} $$

This negative work indicates that gravity opposes the motion of the rock, leading to a decrease in kinetic energy until it reaches its maximum height, where the kinetic energy becomes zero.

In summary, the work done by gravity plays a crucial role in the energy transformations of falling and rising objects, illustrating the fundamental relationship between work, energy, and motion.

You push a 3kg box against a wall for a distance of 2m with a force of 40N that makes a 53° angle with the horizontal, as shown. Calculate the work done by gravity.

• A. 58.8 J
• B. -58.8 J
• C. - 160 J
• D. -80 J

Answer: (B)

Work Done on Curvy (Not straight) Paths

In this scenario, we analyze the work done by gravity on a hiker ascending a mountain from point A at the base to point B at the summit. The hiker does not follow a straight path; instead, they take a winding route with varying speeds and inclinations. To calculate the work done by gravity, we focus on the vertical displacement, denoted as Δy, which in this case is 1,000 meters.

The formula for calculating the work done by gravity is given by:

W_g = -mg Δy

Here, m represents the mass of the hiker, and g is the acceleration due to gravity, approximately 9.8 m/s². It is important to establish a positive direction for our calculations. Since the hiker is moving upward, we define the upward direction as positive, making Δy equal to +1,000 m.

Despite the hiker's irregular path, the work done by gravity is determined solely by the change in height, a concept known as path independence. This means that the work done does not depend on the specific route taken, but only on the initial and final heights.

Substituting the values into the equation, we have:

W_g = -75 kg × 9.8 m/s² × 1,000 m

Calculating this gives:

W_g = -735,000 joules

This negative value indicates that gravity is doing work against the hiker's ascent. Thus, the total work done by gravity during the hike is -735,000 joules, reinforcing the principle that the work done by gravity is dependent only on the change in height, not the path taken.

Here’s what students ask on this topic:

What is the formula for calculating work in physics?

The formula for calculating work (W) in physics is given by:

$W=Fdcos\left(\theta \right)$

where:

• $F$ is the magnitude of the force applied.
• $d$ is the displacement of the object.
• $\theta$ is the angle between the force and the displacement vectors.

This equation shows that work is the product of the force, the displacement, and the cosine of the angle between them.

How do you calculate the work done by gravity?

To calculate the work done by gravity, use the formula:

$W{g}_{}=mg\Delta y$

where:

• $m$ is the mass of the object.
• $g$ is the acceleration due to gravity (9.8 m/s²).
• $\Delta y$ is the vertical displacement.

Gravity does positive work when an object falls and negative work when it rises.

What is the relationship between work and kinetic energy?

The relationship between work and kinetic energy is described by the Work-Energy Theorem, which states that the work done on an object is equal to the change in its kinetic energy:

$W=\Delta \mathrm{KE}$

where:

• $W$ is the work done.
• $\Delta \mathrm{KE}$ is the change in kinetic energy.

Kinetic energy (KE) is given by:

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

where $m$ is the mass and $v$ is the velocity.

How do you determine if work is positive or negative?

Work is positive when the force applied to an object aids its motion, meaning the force and displacement are in the same direction. Conversely, work is negative when the force opposes the object's motion, meaning the force and displacement are in opposite directions. Mathematically, this is determined by the cosine of the angle $\theta$ between the force and displacement vectors:

$W=Fdcos\left(\theta \right)$

If $\theta$ is between 0° and 90°, $cos\left(\theta \right)$ is positive, resulting in positive work. If $\theta$ is between 90° and 180°, $cos\left(\theta \right)$ is negative, resulting in negative work.

What is the significance of the angle in the work formula?

The angle $\theta$ in the work formula $W=Fdcos\left(\theta \right)$ represents the angle between the force vector and the displacement vector. It determines the component of the force that is in the direction of the displacement:

• If $\theta$ = 0°, the force is entirely in the direction of the displacement, and $cos\left(\theta \right)$ = 1, resulting in maximum positive work.
• If $\theta$ = 90°, the force is perpendicular to the displacement, and $cos\left(\theta \right)$ = 0, resulting in zero work.
• If $\theta$ = 180°, the force is opposite to the direction of the displacement, and $cos\left(\theta \right)$ = -1, resulting in maximum negative work.