Work is the energy transfer that happens when an object is moved over a distance by an external force in the direction of the displacement.

Mathematically, work is W = F · x, where F is the applied force and x is the distance moved, that is, displacement. Work is a scalar. The SI unit for work is the joule (J), which is newton‐meter or kg m/s^{2}.

Work corresponds to the direction of the displacement. For example, if a force is applied to a box from an angle, but the box only glides along the horizontal surface, then there is work only in the x direction. There would not be any work in the y direction because there was no displacement in the y direction. Therefore, work can be broken up into the x and y direction:

W_{x} = F_{x} * Δx

W_{y} = F_{y} * Δy

## Work and Potential Energy

The equation for potential energy is

PE,_{gravity} = m * g * Δh

m = mass (kg)

g = acceleration due to gravity = -9.81 m/s^{2}

Δh = change in height = h_{f} – h_{i}

h_{f} = final height

h_{i} = initial height

m * g = F_{g} is the *force of gravity* of the object moving vertically. Δh is the vertical displacement.

Essentially, gravitational potential energy is force * distance, the same equation for work. Therefore, gravitational potential energy is equal to the work needed to move an object from a higher position to a lower position (and vice versa):

PE,_{gravity} = -Work

## Recommended Steps:

Before solving the problem, it is best to draw a diagram of the situation.

1. Identify the force acting on the object and the direction of the force. If the force is applied at an angle, break up the force into its components, keeping in mind the direction of each component.

2. Identify the displacement in each direction. Mark the final and initial position in the x and y directions.

3. Apply variables in the work equation.

## Practice Problems:

1. A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the work that is performed by the action?

Solution:

1. Identify the force acting on the object

The only force acting on the object is the weight of the object, or the force due to gravity. The force is acting downward, so there is no force in the x direction (therefore no work in the x direction):

W = m * g = 3 * 9.81 = 29.43 N

2. Identify the displacement of the object.

The problem indicates that the object is moved from the ground to a ramp that is 0.45 m high. Therefore, the displacement in the y direction is:

Δy = yf – yi = 0.45 – 0 = 0.45 m

3. Apply variables in work equation.

W_{x} = 0 * Δx = 0

W_{y} = F_{y} * Δy = 29.43 * 0.45 = 13.24 J

Alternatively, using the potential energy equation will give us the same result:

PE,_{gravity} = m * g * Δh = 3.0 * 9.81 * 0.45 = 13.24 J

2. A 100 N force is applied to a 10 kg box at 30 degrees from the horizontal on a horizontal plane at a constant speed over a distance of 10 m. Calculate the work generated.

Solution:

1. Identify the force acting on the object

There is a force of 100 N applied at an angle of 30 degrees, which causes the box to move to the right. The gravitational force (weight of object) and normal force also act on the box, but the problem does not indicate a displacement in the y direction, so those forces will not be considered.

Since the applied force is at an angle, it must be broken up into components:

Fx = + 100cos (30) = + 86.60 (force is acting in positive x direction)

Fy = – 100sin (30) = – 50 N (force is acting in negative y direction)

2. Identify the displacement of the object.

The problem indicated that the box moved 10 m to the right. The displacement in the x direction is 10 m.

3. Apply variables in work equation.

W_{x} = F_{x} * Δx = 86.60 * 10 = 866.02 J

**References:**

https://www.cliffsnotes.com/study-guides/physics/classical-mechanics/work-and-energy

https://socratic.org/questions/can-someone-explain-to-me-the-difference-between-potential-and-potential-energy-