Proving that the total mechanical energy is conserved with time

[Hamiltonian]

Homework Statement

Prove that Total mechanical Energy is conserved with time.

Relevant Equations

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To prove: total mechanical energy is constant with time
where $E(t)$ is the total mechanical energy and $V(x(t))$ is the potential energy of the object-system.
$$E(t) = 1/2 mv^2 + V(x(t))$$

taking the the derivative of $E(t)$ with respect time should give 0.
in the third step in the attached file i don't understand why $-f(x(t)) = ma(t)$
also what is the significance of the function $f(x(t))$ which is equal to $-V'(x(t))$

my attempt to a solution:
taking an example of a spring and block system(even though its not very general but I thought I'd get the idea).
so i defined $E = 1/2mv^2 + 1/2kx^2$
from this i got $E' = v(ma + kx)$
I could not further simplify this.

also I think I was able to prove the above using the work energy theorem $\Delta KE = -\Delta PE$ hence $\Delta(KE + PE) = 0$
so what am missing and doing wrong in the above method stated by me.

 

[BvU]

It's called Newton's law, usually written as $F = ma$

In your example $F = - kx$ for a spring.

 

[etotheipi]

If you like, you can re-state their proof from a slightly different starting point; assuming the particle moves in some potential $V(x)$, then$$m\ddot{x} = F_x = -V'(x)$$ $$m\ddot{x}\dot{x} = - V'(x) \dot{x}$$ $$\frac{d}{dt} \left(\frac{1}{2}m\dot{x}^2 \right) = -\frac{d}{dt} V(x)$$ $$\frac{d}{dt} \left( \frac{1}{2}m\dot{x}^2 + V(x) \right) = 0 \implies \frac{1}{2}m\dot{x}^2 + V(x) = E = \text{constant}$$