Calculating the Work Done by a Man Lifting a 70 kg Weight 5 Meters Vertically

Introduction:

In physics, work is defined as the amount of energy transferred to or from an object via the application of force along a displacement. In this article, we will explore the concept of work done by a man who lifts a 70 kg weight vertically 5 meters. We will break down the physics involved and provide a step-by-step solution to this real-world problem.

Understanding the Concept of Work in Physics

Work (W) is calculated as the product of force (F) and displacement (d) in the direction of the force. This can be expressed by the equation:

W F × d

However, if the displacement is horizontal and the force is vertical, no work is done. Conversely, if the displacement is vertical and the force is also vertical, work can be calculated with the above equation.

Breaking Down the Problem

The problem states that a man is holding a 70 kg weight and moving it 7 meters horizontally and 5 meters vertically. Let's analyze this situation step by step:

Vertical Movement: When the man is vertically lifting the weight 5 meters, it represents the condition where the work is done. This is because in this segment, the force applied (by the man) is in the same direction as the displacement (upwards).

Horizontal Movement: When the man moves the weight horizontally 7 meters, no work is done. This is because the force exerted is horizontal, but the displacement is also horizontal, meaning there is no component of force in the direction of displacement.

Calculating the Force

To calculate the force an individual must exert to lift a 70 kg mass, we use the formula for weight:

F m × g

Where:

m mass (70 kg)

g acceleration due to gravity (9.8 m/s2)

Plugging these values into the equation:

F 70 kg × 9.8 m/s2 686 N (Newtons)

Calculating the Work Done

Now that we have the force required to lift the weight, we can calculate the work done during the vertical movement. Since the weight is lifted 5 meters vertically, we can calculate the work done using the equation:

W F × d

Here:

F 686 N (Newtons)

d 5 m (meters)

Therefore:

W 686 N × 5 m 3430 N·m or 3430 Joules

Thus, 3430 Joules of work is done.

Conclusion

Understanding the concept of work in physics allows you to solve practical problems such as this one. Remember, work is only done when the force applied is in the same direction as the displacement. In this example, when the man is lifting a 70 kg weight vertically 5 meters, the work done is 3430 Joules.

Grasping these fundamental principles can help you in various fields, from engineering to sports science. Understanding how to apply these formulas can make problem-solving easier in everyday scenarios.

Additional Resources

For more in-depth learning, we recommend exploring these topics further:

Work and Energy: A comprehensive guide to Work, Kinetic Energy, and Potential Energy.

Application of Gravity in Physics: Understanding the force of gravity and its role in everyday physics problems.

Mechanics in Physics: An overview of mechanics, which deals with the motion of bodies and the forces acting on them.