Solve Spring Force Problem: V at x=4 & x for V=5

[PrudensOptimus]

The only force acting on a 2.0kg body as it moves along the positive x-axis has an x compenent Fx = -6x Newtons, where x is in meters. The velocity of the body at x = 3.0m is 8.0m/s.


(a). What is the velocity of the body at x = 4.0m?
(b). At what positive value of x wil the body have a velocity of 5.0m/s?

(a). Since K = W = Integral ( F dx )

and F = Fx = -6x

K = -3 (x ^2) evaluate from 0 to 4
= 48
m/2 v^2 = 48
v = 6.9m/s

correct?

(b). since m/2 v^2 = -3x^2
and v = 5

x = sqrt ( -25/3 )

is not right because it would give me a complex number...


Please advise!

 

[daniel_i_l]

In this case the body had an initial speed at x=0 (otherwise how could it have a positive velocity at x=3)
Find the initial velocity using the velocity at x=3, then you should get the correct answers for a and b.

 

[andrevdh]

Have you done SHM, because this is such type of problem? You can therefore use the SHM equations (or try to derive them yourself!). Note that since
F=-6x
we have that
a=-\frac{6}{m}x
which gives
a=-\omega^2x
Not to confuse you, but as the problem is stated one might think that the force also has an y-component (which I think is not the case, since then it is not possible to answer the questions).

 

[Tide]

HINT: (a) The work done represents the change in kinetic energy.

HINT: (b) The change in potential plus kinetic energy is 0.

 

[PrudensOptimus]

so the sum of all external forces = 0?

 

[Tide]

Energy is conserved so

m v^2 + k x^2 = m v_0^2 + k x_0^2[/itex]

 

[andrevdh]

so the sum of all external forces = 0?

No, but the work done by the conservative restoring force of the spring is included in the work-energy equation, so we do not need to consider the work done by this force if we include it as the potential energy of the system (the mass and the spring) - which is given by \frac{1}{2}kx^2 as indicated by Tide.

 

[PrudensOptimus]

Could you explain what are the differences between sum of all forces = 0 and K + U = K + U?

 

[andrevdh]

From the work kinetic energy theorem we have that the change in kinetic energy of a body is given by the work done by all of the forces that acts on it. If there are no forces, or the forces do not do work on the body, the kinetic energy of the body stays constant. We can exlude the work done by a conservative force from the work side of this calculation if we include the potential energy of the system (which is the negative of the work done by the conservative force therefore) on the energy side of the equation. If no other force acts on the body the total energy of the body will then remain constant or
K_1+U_1=K_2+U_2