# Trouble understanding total mechanical energy and the second law of Newton

• studentsuff
In summary, the relevant equations for constant acceleration and work conservation are equated to each other and multiplied by mass to account for gravity. This results in the equivalent of the energy equation and produces the same answer.
studentsuff
Homework Statement
'Suppose your hand moves upward by 0.50m while you are throwing the ball. The ball leaves your hand with an upward velocity of 20.0 m/s. Find the magnitude of the force (assumed constant) that your hand exerts on the ball. Ignore air resistance.' Now my question is not 'What is the magnitude' but rather: why did I get (roughly) the same answer using F=ma when you were supposed to use the total mechanical energy (W+K1+U1=K2+U2). So I'm more confused about how the 2 formulas are 'related', what the force actually represents in both, when to use what and if there is an actual difference.
Relevant Equations
a = (((v2)^2)-((v2)^2))/(2(x2-x1))
F = m*a
W(other) + K1 + U1 = K2 + U2
F = (K2 + U2 - K1 - U1) / s
Question

studentsuff said:
Relevant Equations:: a = (((v2)^2)-((v2)^2))/(2(x2-x1))
Since the force is constant so is the acceleration. The above equation for constant acceleration effectively is the work conservation equation; multiply it out to get rid of the division:
##a \Delta x=\frac 12( v_2^2-v_1^2)##
Then multiply by mass:
##ma \Delta x= \frac 12( mv_2^2-mv_1^2)##
Or
##F \Delta x= \frac 12( mv_2^2-mv_1^2)##

Lnewqban and studentsuff
When using "F = ma", you have to realize that ma is equal to the net force acting on the ball. Since the ball is subject to gravity, to get an acceleration a you have to apply a force F such that
F - mg = ma
F = ma + mg
When you take account of gravity in this way, you find that it is equivalent to accounting for the gravitational potential energy (U2 and U1) in the energy equation. Then you should find you get exactly the same answer (not roughly) using both approaches.

## 1. What is total mechanical energy?

Total mechanical energy is the sum of an object's kinetic energy and potential energy. Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object has due to its position or state.

## 2. How is total mechanical energy related to the second law of Newton?

The second law of Newton states that the net force acting on an object is equal to the object's mass multiplied by its acceleration. Total mechanical energy is related to this law because it is a measure of an object's energy, which is affected by the forces acting on it.

## 3. Why is it important to understand total mechanical energy?

Understanding total mechanical energy is important because it allows us to predict the behavior of objects and systems. It also helps us understand the relationship between energy and forces, which is essential in many areas of science and engineering.

## 4. How can I calculate total mechanical energy?

To calculate total mechanical energy, you need to know the object's mass, velocity, and position. The formula for total mechanical energy is E = KE + PE, where E is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy.

## 5. What happens to total mechanical energy when an object is in motion?

When an object is in motion, its total mechanical energy remains constant as long as there are no external forces acting on it. This is known as the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed.

• Introductory Physics Homework Help
Replies
9
Views
318
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
13
Views
400
• Introductory Physics Homework Help
Replies
5
Views
653
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
453
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
186
• Introductory Physics Homework Help
Replies
1
Views
526