Respuesta :
Answer:
Explanation:
Given equation is ,
x =t + 2 t³ ,
dx/dt = velocity ( v ) = 1 + 6 t²
a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²
b ) Acceleration = dv /dt = 12 t .
force( F ) = mass x acceleration = 4 x 12 t = 48 t
Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )
work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; =  [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² =  864 J
Answer:
Givens:
- [tex]x=t+2.0t^{3}[/tex]
- [tex]m=4.00 \ kg[/tex]
We know that kinetic energy is:
[tex]K=\frac{1}{2}mv^{2}[/tex]
So, we just need to calculate the speed. We have the equation of the movement, if we derivate that expression, we'll have the speed:
[tex]\frac{dx}{dt}=v\\ v=\frac{d}{dt}(t+2.0t^{3})\\v=1+2(3)t^{2} \\v=6t^{2}+1[/tex]
Which is the speed at any time t.
Now, we replace the expression to find the kinetic energy at any time t:
[tex]K=\frac{1}{2}m(6t^{2}+1)^{2}[/tex]
[tex]K=\frac{1}{2}(4)(36t^{4}+12t^{2}+1)\\K=2(36t^{4}+12t^{2}+1)\\K=72t^{4}+24t^{2}+2[/tex]
So, this is the kinetic energy at energy at any time t.
Through derivation we can find the acceleration and force at any time t:
[tex]a=\frac{dv}{dt}\\a=\frac{d}{dt}(6t^{2}+1)\\a=12t[/tex]
Also, we know that [tex]F=ma[/tex]
Replacing values: [tex]F=4(12t)=48t[/tex]
The power is define as the product of the force and the velocity:
[tex]P=Fv=48t(6t^{2}+1)=288t^{3}+48[/tex]
At last, we now that the work is define as: [tex]W=\int Fdx[/tex]
So, we just replace the force, and integrate it between t=0 and t=2 sec.
[tex]W=\int _{0}^{2} 48t(6t^{2}+1)dt\\W=\int_{0}^{2} (288t^{3}+48t)dt\\W= \frac{288t^{4} }{4}+\frac{48t^{2}}{2} ]_{0}^{2} \\W= 72t^{4}+24t^{2}]_{0}^{2}\\W=72(2)^{4}+24(2)^{2}\\W= 1152 + 96=1248 \ J[/tex]
