Work = Force \times Distance

Where:

**Work**: the amount of energy transferred to or from an object by means of a force acting on the object**Force**: a push or pull upon an object resulting from the object’s interaction with another object**Distance**: the amount of space between two points or locations.

In physics, work is defined as the product of the force applied to an object and the distance covered by that object in the direction of the force. Mathematically, this can be expressed as:

W = F \cdot d

Where:

- W represents the work done,
- F represents the force applied, and
- d represents the distance covered.

One of the classic experiments that demonstrates the relationship between work, force, and distance covered is the lifting of a weight against gravity. In this experiment:

- The force applied is the weight of the object being lifted.

- The distance covered is the vertical distance the object is lifted.

- The work done is equal to the force applied times the distance covered.

For example, if a person lifts a 10 kg weight vertically upwards by 2 meters, the work done can be calculated as follows:

- Force applied (weight) = mass x gravity = 10 kg x 9.8 m/s^2 = 98 N

- Work done = Force x Distance covered = 98 N x 2 m = 196 J

The concept of work is widely used in various fields, including physics, engineering, and everyday activities. Some practical applications include:

- Calculating the work done by a machine to move an object from one point to another.
- Determining the amount of work required to lift an object to a certain height in construction projects.
- Understanding the energy transfer involved in lifting weights during workout sessions.

Overall, the equation Work = Force x Distance Covered provides a fundamental understanding of the relationship between force, distance, and work done in various physical scenarios. By applying this equation, scientists and engineers can accurately calculate and predict the amount of work required for a given task.

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Work

**1. A force of 50 N is applied to an object, causing it to move a distance of 10 meters. Calculate the work done on the object.**

\text{Given:} \ \text{Force} = 50 \ N, \ \text{Distance} = 10 \ m

\text{Work} = \text{Force} \times \text{Distance}

\text{Work} = 50 \ N \times 10 \ m

\text{Work} = 500 \ J

Therefore, the work done on the object is 500 Joules.

**2. If a force of 30 N is applied to lift a box vertically upwards by a distance of 5 meters, calculate the work done.**

\text{Given:} \ \text{Force} = 30 \ N, \ \text{Distance} = 5 \ m

\text{Work} = \text{Force} \times \text{Distance}

\text{Work} = 30 \ N \times 5 \ m

\text{Work} = 150 \ J

The work done to lift the box vertically upwards is 150 Joules.

**3. A force of 80 N is used to push a heavy block horizontally by a distance of 2 meters. Calculate the work done in moving the block.**

\text{Given:} \ \text{Force} = 80 \ N, \ \text{Distance} = 2 \ m

\text{Work} = \text{Force} \times \text{Distance}

\text{Work} = 80 \ N \times 2 \ m

\text{Work} = 160 \ J

Hence, the work done in moving the block horizontally is 160 Joules.

**4. An object is pulled with a force of 100 N along a rough surface, covering a distance of 3 meters. Find the work done on the object.**

\text{Given:} \ \text{Force} = 100 \ N, \ \text{Distance} = 3 \ m

\text{Work} = \text{Force} \times \text{Distance}

\text{Work} = 100 \ N \times 3 \ m

\text{Work} = 300 \ J

Thus, the work done on the object when pulled along the rough surface is 300 Joules.

**5. Calculate the work done when an applied force of 120 N moves an object through a distance of 4 meters at an angle of 30 degrees with the horizontal.**

\text{Given:} \ \text{Force} = 120 \ N, \ \text{Distance} = 4 \ m, \ \text{Angle} = 30^\circ

\text{Work} = \text{Force} \times \text{Distance} \times \cos(30^\circ)

\text{Work} = 120 \ N \times 4 \ m \times \cos(30^\circ)

\text{Work} = 480 \cos(30^\circ) \ J

\cos(30^\circ) = \frac{\sqrt{3}}{2}

\text{Work} = 480 \times \frac{\sqrt{3}}{2} \ J

\text{Work} = 240 \sqrt{3} \ J

Hence, the work done on the object at an angle of 30 degrees with the horizontal is \( 240 \sqrt{3} \) Joules.

More Worked Examples on Work
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**1. Calculate the work done when a force of 20 N is exerted over a distance of 10 m.**

Work = Force \times Distance

Work = 20 N \times 10 m = 200 J

**2. A person lifts a 5 kg box up a flight of stairs, exerting a force of 50 N over a vertical distance of 3 m. Calculate the work done.**

Work = Force \times Distance

Work = 50 N \times 3 m = 150 J

**3. A car engine exerts a force of 5000 N to move a car 100 m. Calculate the work done by the engine.**

Work = Force \times Distance

Work = 5000 N \times 100 m = 500,000 J

**4. A child pushes a 10 kg shopping cart with a force of 30 N over a distance of 5 m. Calculate the work done.**

Work = Force \times Distance

Work = 30 N \times 5 m = 150 J

**5. A student pushes a sled with a force of 40 N over a distance of 8 m. Calculate the work done.**

Work = Force \times Distance

Work = 40 N \times 8 m = 320 J

**6. Calculate the work done when a force of 15 N is applied to move an object a distance of 6 m.**

Work = Force \times Distance

Work = 15 N \times 6 m = 90 J

**7. A crane lifts a load with a force of 2000 N over a vertical distance of 20 m. Calculate the work done by the crane.**

Work = Force \times Distance

Work = 2000 N \times 20 m = 40,000 J

**8. A spring is compressed by a force of 100 N. If the spring constant is 50 N/m, calculate the amount of work done in compressing the spring by 0.2 m.**

Work = \frac{1}{2} k x^2

Work = \frac{1}{2} \times 50 N/m \times (0.2 m)^2 = 1 J

**9. Calculate the work done when a 30 N force is used to push a cart up a ramp inclined at 30 degrees over a distance of 10 m.**

Work = Force \times Distance \times \cos(\theta)

Work = 30 N \times 10 m \times \cos(30^\circ) = 259.8 J

**10. An object is pulled with a force of 25 N at an angle of 60 degrees to the horizontal, covering a distance of 6 m. Calculate the work done.**

Work = Force \times Distance \times \cos(\theta)

Work = 25 N \times 6 m \times \cos(60^\circ) = 75 J

**11. A machine applies a force of 150 N vertically to lift a load 5 m. Calculate the work done.**

Work = Force \times Distance

Work = 150 N \times 5 m = 750 J

**12. Calculate the work done by a force of 40 N acting on an object moving horizontally for a distance of 12 m.**

Work = Force \times Distance \times \cos(\theta)

Work = 40 N \times 12 m \times \cos(0^\circ) = 480 J

**13. A sled is pulled with a force of 60 N at an angle of 45 degrees to the horizontal, covering a distance of 8 m. Calculate the work done.**

Work = Force \times Distance \times \cos(\theta)

Work = 60 N \times 8 m \times \cos(45^\circ) = 339.41 J

**14. A weightlifter lifts a 100 kg barbell vertically with a force of 980 N over a distance of 2 m. Calculate the work done.**

Work = Force \times Distance

Work = 980 N \times 2 m = 1960 J

**15. Calculate the work done by a force of 35 N acting horizontally on an object moving up an inclined plane with a height of 4 m and a length of 6 m.**

Work = Force \times Distance \times \cos(\theta)

Work = 35 N \times 6 m \times \cos(\arctan(4/6)) = 210 J

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