a mehanics / physics question.?

Question

an object has masss =3kg & is moving along x-axis with v=2m/s as it passes troguh x=0. it is subjected to a single force Fx tht varies with position. (i) what is kinetic energy as object passes through x=0? (ii) what is the work done by the force as the particle moves from x=0 to x=4. (iii) what is the speed of particle when it is at x=4m?
for (i) i got 6 J and (ii) i got 24 which i'm not sure about. i'm not sure how u work out the speed from work or th kinetic energy

Answer 1

To go beyond the first part we need to know in precise terms how the force varies with distance.

Answer 2

i) Kinetic energy = 1/2mv^2
m = 3, v = 2, 1/2mv^2 = 1/2 x 3 x 4 = 6 Joules.

ii) Work done = force x distance moved.
Force = fx, distance = 4 m
There is a linear increase in force with distance 'x'. so, take the average force between x = 0 and x = 4:
[Fx(0) + Fx(4)]/2 = 2F

So, the average force (2F) x distance (4) = 8F.
For a force Fx, of course, you need the value of 'F', in order to get the answer. You cannot derive this value from the first part of the question, as it only gives you data for the state Fx(0), where 0 force is acting.

iii) The total energy of particle at x = 4m, is the energy at
Fx(0) + the work done:
6 + 8F. This again, = 1/2mv^2. So, rearranging:
v= ((1/2m)^1/2)/(6 + 8F). Substitute in the mass, gives v = 1.22/(6 + 8F). You cannot solve further without knowing 'F'

Answer 3

Kinetic energy (x=0)= 1/2M*V(x=0)^2
Work done W total= Ftotal * d = Integration F(x) * dx (x=0,....... X=4, increment = 4 = d)
Force (Ftotal) = Integration F(x) (We have to know how force varies with distance x, we don't know the function that relates both entities)
Kinetic energy (X=4) = Kinetic Energy (X=0) + Work done (W) = 1/2M*V(x=4)^2
Summarizing; you need the Force F(x) function to calculate it.
We assume that there is no friction.