A particle (mass = 0.41 kg) moves in one dimension (x) and the potential energy

Question

A particle (mass = 0.41 kg) moves in one dimension (x) and the potential energy is V(x) = (?1/2) q x2 with q = 6.12 J/m2. The initial conditions at t = 0 are velocity = 0 and position x0 = 2.2 m.
(A) What is the force?
(A) What is the force?
v(x) = -(1/2)qx^2
F(x) = dv/dx = qx
f(x) = 6.12x

At x=x0=2.2,
F(x)=6.12(x0) = 6.12*2.2= 13.46 J/m


I don't have the other two parts to the question...

(B) Solve for the motion of the particle, from conservation of energy. x(t) has the form A cosh(?t). Calculate the value of ?.


(C) Calculate the time t when x = 2 x0.

Explanation / Answer

F = qx So a = 14.93x So d2x/ dt2 = 14.93x Which By solving will come in the form ofx= Acosh(Bt) Use initial cond to find A and B After finding equation u wil also be able to solve c part

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