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

• studentsuff

#### 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

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.