A particle of mass 2.5 kg moves in a conservative force field. Its potential energy curve is shown below. From the curve, determine (a) the total mechanical energy of the particle at x = 1.3 m if...
A particle of mass 2.5 kg moves in a conservative force field. Its potential energy curve is shown below. From the curve, determine (a) the total mechanical energy of the particle at x = 1.3 m if it has a speed of 8 ms−1 and (b) the minimum total energy at which the particle can escape from the force field. If Etot = 150 J, where would the particle have (i) zero velocity, (ii) maximum velocity?
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).`
The second part of the question:
If the total energy of the particle is `E_t = 150 J` , then the potential energy of the particle E(x) is either less or equal to 150 J, as follows from the equation
`E(x) + mV^2/2 = E_t` .
This corresponds to the region on the potential energy curve which lies below the horizontal line corresponding to E(x) = 150 J. This region on the interval where x is greater than 1.3 m and less than 3.4 m: `1.3 <=x<=3.4` . So, if the total energy of the particle is 150 J, the particle will be "trapped" by the force field in between these two values of x.
The velocity of the particle will be zero when `E(x) = E_t` , which happens at x = 1.3 m and x = 3.4 m.
The velocity will be maximum at the point where E(x) is minimum, which is at the point x = 2.6 m. At this point, the particle is in equilibrium.