Question

The following position-dependent net force acts on a 3 kg block:

[F subscript n e t end subscript open parentheses x close parentheses equals open parentheses 3 straight N over straight m squared close parentheses x squared]If the block starts at rest at x = 2 m, what is the magnitude of its linear momentum (in kgm/s) at x = 4 m?

Answer 1

`p=m.v=37\ kg.m.s^(-1)`

Step-by-step explanation:

Given:

• 
  `F_(net)=3x^2\ [N]`

• The initial position of the block,
  `x=2\ m`

• mass of the block,
  `m=3\ kg`

• final position of the block,
  `x=4\ m`

WE know from the Newton's second law:

`(d)/(dt) (p)=F`

where:

`p=` momentum

`F=` force

`t=` times

Now put the values

`(d)/(dt) (mv)=3\cdot x^2`

`m.(d)/(dt) (v)=4* x^2`

Now integrate both sides from final limit to initial:

`m.v=\int\limits^4_2 {3x^2} \, dx`

`m.v=[(3x^3)/(3) ]^4_2`

`m.v=4^3-2^3`

`p=m.v=56\ kg.m.s^(-1)`

Answer 2

Step-by-step explanation:

Given

F_net(x) = (3 x²) N

m v dv / dt = 3 x²

m v dv = 3 x² dx

integrating on both sides and taking limit from x = 2 to 4 m

m v² / 2 - 0 = 3 x³ / 3

mv² / 2 = 4³ - 2³

mv² / 2 = 64 - 8

3 x v² /2 = 56

v = 6.11 m / s

linear momentum

= m v

= 3 x 6.11

= 18.33 kgm/s