Respuesta :
Answer:
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
Answer:
[tex]p=m.v=37\ kg.m.s^{-1}[/tex]
Explanation:
Given:
- [tex]F_{net}=3x^2\ [N][/tex]
- The initial position of the block, [tex]x=2\ m[/tex]
- mass of the block, [tex]m=3\ kg[/tex]
- final position of the block, [tex]x=4\ m[/tex]
WE know from the Newton's second law:
[tex]\frac{d}{dt} (p)=F[/tex]
where:
[tex]p=[/tex] momentum
[tex]F=[/tex] force
[tex]t=[/tex] times
Now put the values
[tex]\frac{d}{dt} (mv)=3\cdot x^2[/tex]
[tex]m.\frac{d}{dt} (v)=4\times x^2[/tex]
Now integrate both sides from final limit to initial:
[tex]m.v=\int\limits^4_2 {3x^2} \, dx[/tex]
[tex]m.v=[\frac{3x^3}{3} ]^4_2[/tex]
[tex]m.v=4^3-2^3[/tex]
[tex]p=m.v=56\ kg.m.s^{-1}[/tex]
