Calculating Work Done in Lifting a Weight: A Student's Squat and Joules of Effort

In the realm of physics, the concept of work and energy plays a crucial role in understanding the mechanical interactions between objects. This article delves into the precise calculation of the work done when a student lifts a 50-pound weight approximately 4 feet in a span of 5 seconds, using the unit of joules and the fundamental formula for work.

The Physics Behind the Lift

The problem at hand involves understanding the relationship between force, distance, and the angle at which the force is applied. The formula for work is given by Work Force x Distance x cosθ, where θ is the angle between the force and the direction of motion. In this scenario, the student lifts the ball vertically, making the angle θ equal to 0 degrees, thus simplifying the equation to Work Force x Distance.

Conversion of Units

To calculate the work done, the first step is to convert the given units into consistent units. We know that 1 pound (lb) is approximately equal to 4.45 Newtons (N), and 1 foot (ft) is approximately equal to 0.3048 meters (m).

Converting Weight to Force in Newtons

The weight of the ball is 50 pounds. Using the conversion factor, we can determine the force in Newtons:

222.41 N 50 lb x 4.44822 N/lb

Converting Distance to Meters

The height the ball is lifted is 4 feet. To convert this distance into meters:

1.2192 m 4 ft x 0.3048 m/ft

Calculating the Work Done

Using the formula for work, we can now calculate the work done by the student:

Work 222.41 N x 1.2192 m x 1 271.74 Joules

This means that the student has completed approximately 271.74 Joules of work.

Considering the Work-Energy Theorem

Another way to approach this problem is through the Work-Energy Theorem, which states that the work done on an object is equal to the change in its kinetic and potential energy. The ball initially possessed only kinetic energy as it was on the ground. When lifted to a height of 4 feet, this kinetic energy is converted entirely into potential energy.

Calculating Potential Energy at the Highest Point

The mass of the ball can be calculated as 50 pounds, which is approximately 22.68 kilograms.

PE mgh 22.68 kg x 9.81 m/s2 x 1.219 m 271.2163 Joules

Thus, the potential energy at the highest point is 271.2163 Joules. This energy is equivalent to the kinetic energy of the ball at the ground level and represents the work done by the student lifting the ball to a height of 4 feet.

Conclusion

This detailed analysis shows that the work done in lifting the ball can be calculated through both the direct application of the work formula and the Work-Energy Theorem. Understanding these principles not only helps in solving specific physics problems but also provides insights into the fundamental laws governing mechanical interactions. By accurately converting units and applying the appropriate formulas, we can quantitatively describe the effort and energy involved in lifting a weight.