Calculating Net Work Done on an Accelerating Object: A Detailed Guide
Understanding the net work done on an accelerating object is crucial for anyone studying physics or engineering. In this article, we will walk through the steps to calculate the net work done when a person walks 2 meters with an acceleration of 5 m/s2, holding an object of mass 2 kg. We will use the work-energy principle, kinematics, and some basic physics principles to derive the solution.
Understanding the Problem
The problem at hand is: A person walks 2 meters with an acceleration of 5 m/s2, holding an object of mass 2 kg. We need to calculate the net work done on the object.
Step-by-Step Solution
1. Determine the Final Velocity
To find the net work done, we first need to calculate the final velocity of the object after it travels 2 meters with an acceleration of 5 m/s2. We can use the kinematic equation for this purpose:
v^2 u^2 2as
Where:
v final velocity u initial velocity (0 m/s, since the object starts from rest) a acceleration (5 m/s2) s distance traveled (2 meters)Plugging in the values, we get:
v^2 0 2 times 5 times 2
v^2 20
v sqrt{20} approx 4.47 , m/s
2. Calculate the Change in Kinetic Energy (ΔKE)
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. We can calculate this change using the following formula:
ΔKE frac{1}{2} m v^2 - frac{1}{2} m u^2
Where:
ΔKE change in kinetic energy m mass of the object (2 kg) v final velocity (4.47 m/s) u initial velocity (0 m/s)Plugging in the values, we get:
ΔKE frac{1}{2} times 2 times (4.47^2) - frac{1}{2} times 2 times 0^2
ΔKE 1 times 20 approx 20 , J
Thus, the net work done on the object is approximately 20 Joules.
Alternative Method: Using Force and Displacement
An alternative approach to solving this problem directly involves using the formula for work:
W F times d m times a times d
Where:
W work done F force (2 kg times 5 m/s2) d displacement (2 meters)Plugging in the values, we get:
W 2 , kg times 5 , m/s^2 times 2 , m 20 , J
Both methods yield the same result: the net work done on the object is 20 Joules.
It's worth noting that the weight of the object (due to gravity) does not perform any work on the object when the object is moved horizontally because the force of gravity is perpendicular to the displacement. Therefore, only the component of the force parallel to the displacement (in this case, the force due to the person's acceleration) contributes to the work done.
Conclusion
Understanding the concepts of work, kinetic energy, and force can help solve more complex problems in physics. This detailed approach and the problem-solving techniques shown here can be applied to various scenarios involving motion and work.
If you need more help with physics problems or want to explore similar topics in more depth, consider checking out the RCM Science Channel on YouTube, where detailed problem-solving videos are available.
Credit: Photos, diagrams, and additional resources have been appropriately sourced and used for educational purposes only. Further details, relevant links, and additional references are always encouraged for a comprehensive understanding.