Hello! I'll answer the first part of your question.
Total energy of a particle is kinetic energy `(mV^2)/2` plus potential energy `E.` The force field is conservative so the potential energy depends only on the position `x` of a particle. Total energy of a particle has the same value for any `x` because of energy conservation law.
a) at `x=1.3 m` the potential energy is `150 J` as we see from the graph. The...
Hello! I'll answer the first part of your question.
Total energy of a particle is kinetic energy `(mV^2)/2` plus potential energy `E.` The force field is conservative so the potential energy depends only on the position `x` of a particle. Total energy of a particle has the same value for any `x` because of energy conservation law.
a) at `x=1.3 m` the potential energy is `150 J` as we see from the graph. The kinetic energy is `(2.5*8^2)/2=80 (J),` so the full energy is `150 J + 80 J = 230 J.`
b) a particle escapes a force field if its `x` can be arbitrary large. Denote the total particle's energy as `E_t` and consider the equation `(mV^2)/2+E(x)=E_t.`
From it we obtain `V^2(x)=2/m (E_t-E(x))gt=0.` From the graph we see that `E(x)->250` as `x->+oo.` Also recall that the displacement is the integral of the velocity (the velocity is the derivative of the displacement with respect to time).
b1) if `E_tlt250 J,` then `x` cannot be arbitrary large (`V^2=E_t-E(x)` will become negative which is impossible), and a particle won't escape.
b2) if `E_tgt250 J,` then `V(x)gt=sqrt(2/m (E_t-250))=Cgt0.` Therefore
`x(t)=int_0^t V(x(t)) dtgt=C*t->+oo.` So a particle will escape.
b3) if `E_t=250 J,` then the answer may be different for different `E(x).`
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