Calculating Work Done: Force and Displacement

The question asks us to find the work done when a specific force pushes a cart over a certain distance. Work done in physics is a measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. The mass of the cart is provided (30 kg), but in this particular scenario, it is not needed to calculate the work done, as we are directly given the force applied and the displacement.

Understanding Work Done in Physics

Work done (\(W\)) is defined as the product of the force (\(F\)) applied on an object and the displacement (\(d\)) of the object in the direction of the force. Mathematically, this is expressed as:

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

Where:

• \(W\) is the work done
• \(F\) is the magnitude of the force
• \(d\) is the magnitude of the displacement
• \(\theta\) is the angle between the force vector and the displacement vector

In this problem, the force of 750 N pushes the cart by 16 m. We can assume that the force is applied in the direction of the displacement, which means the angle \(\theta\) between the force and displacement is 0 degrees. The cosine of 0 degrees is 1 (\(\cos(0^\circ) = 1\)).

Therefore, the formula for work done simplifies to:

\(W = F \times d\)

Step-by-Step Work Done Calculation

Let's use the given values to calculate the work done:

• Force (\(F\)) = 750 N
• Displacement (\(d\)) = 16 m

Using the simplified formula \(W = F \times d\):

\(W = 750 \, \text{N} \times 16 \, \text{m}\)

\(W = 12000 \, \text{J}\)

The work done is calculated in Joules (J), as the force is in Newtons (N) and displacement is in meters (m). The standard unit of work (and energy) in the International System of Units (SI) is the Joule.

Converting Work Done from Joules to Kilojoules

The question asks for the work done in kilojoules (kJ). One kilojoule is equal to 1000 Joules. To convert Joules to Kilojoules, we divide the value in Joules by 1000.

\(1 \, \text{kJ} = 1000 \, \text{J}\)

So, to convert 12000 J to kJ:

\(W \, (\text{kJ}) = \frac{12000 \, \text{J}}{1000 \, \text{J/kJ}}\)

\(W \, (\text{kJ}) = 12 \, \text{kJ}\)

Thus, the work done is 12 kJ.

Summary of Work Done Calculation

Here is a summary of the values and the calculation steps:

Calculation:

• Work Done (\(W\)) in Joules = \(F \times d = 750 \, \text{N} \times 16 \, \text{m} = 12000 \, \text{J}\)
• Work Done (\(W\)) in Kilojoules = \(\frac{12000 \, \text{J}}{1000} = 12 \, \text{kJ}\)

Final Answer on Work Done

The work done by the force of 750 N pushing the cart by 16 m is 12 kJ.

Revision Table: Work, Force, and Displacement

Additional Information: Work and Energy Concepts

Work done is closely related to energy. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. When positive work is done on an object, its kinetic energy increases (assuming no other energy transfers). When negative work is done, its kinetic energy decreases.

Different types of forces can do work. For example, friction can do negative work (opposing motion), gravity can do work (positive or negative depending on movement direction), and applied forces like in this problem can do work.

Understanding the angle \(\theta\) is crucial:

• If \(\theta = 0^\circ\), force and displacement are in the same direction, \(\cos(0^\circ) = 1\), \(W = F \times d\) (Maximum positive work).
• If \(\theta = 90^\circ\), force is perpendicular to displacement, \(\cos(90^\circ) = 0\), \(W = 0\) (No work done by this force).
• If \(\theta = 180^\circ\), force and displacement are in opposite directions, \(\cos(180^\circ) = -1\), \(W = -F \times d\) (Maximum negative work).

In this problem, the applied force is assumed to be doing positive work on the cart, increasing its kinetic energy or overcoming opposing forces like friction, although friction is not mentioned in the problem statement. The question only asks for the work done by the specific 750 N force over the given displacement.