Respuesta :
Answer:
8.31 m/s²
567.32 N
Explanation:
The equation is given as
y = 0.2 x²
taking derivative both side relative to "x"
[tex]\frac{dy}{dx}[/tex] = 0.2 x eq-1
at x = 8 m
[tex]\frac{dy}{dx}[/tex] = 0.2 x 8 = 1.6
m = slope = [tex]\frac{dy}{dx}[/tex] = 1.6
tanθ = 1.6
θ = 58 deg
Taking derivative both side relative to "x"
[tex]\frac{d^{2}y}{dx^{2}}[/tex] = 0.2
Radius of curvature is given as
[tex]r = \frac{(1 + m^{2})^{\frac{3}{2}}}{(\frac{d^{2}y}{dx^{2}})}[/tex]
[tex]r = \frac{(1 + 1.6^{2})^{\frac{3}{2}}}{(0.2)}[/tex]
[tex]r = 33.6[/tex]
rate of increase of speed is given as
[tex]a_{t}[/tex] = g Sinθ
[tex]a_{t}[/tex] = (9.8) Sin58
[tex]a_{t}[/tex] = 8.31 m/s²
centripetal acceleration is given as
[tex]a_{c} = \frac{v^{2}}{r}[/tex]
[tex]a_{c} = \frac{4^{2}}{33.6}[/tex]
[tex]a_{c} = 0.48[/tex]
F = normal force
Force equation is given as
F - mg Cosθ = m [tex]a_{c}[/tex]
F - (100) (9.8) Cos58 = (100) (0.48)
F = 567.32 N
Answer:
a) Rate of increase in speed=[tex]a_{t}=9.361[/tex] [tex]m/s^{2}[/tex]
b) Normal force =[tex]N=310.34[/tex] [tex]N[/tex]
Explanation:
Given slope=[tex]y=0.2*x^{2}[/tex]
Taking derivatives of both sides with respect to x,we have
[tex](dy/dx)=(d/dx)(0.2x^{2})[/tex]
[tex](dy/dx)=0.4*x[/tex]
at [tex]x=8[/tex]
[tex](dy/dx)=3.2[/tex]
[tex](d^{2}y/ d^{2} x)=0.4[/tex]
Where this will give us the tanΘ.
tanΘ=3.2
Θ=72.6°
Now applying equations of equilibrium in tangential direction
∑[tex]F_{t}=m*a_{t}[/tex]
where ∑[tex]F_{t}=m*g*Sin[/tex]Θ
where [tex]a_{t}=g*Sin[/tex]Θ
[tex]g=9.8[/tex] [tex]m/s^{2}[/tex]
Rate of increase in speed=[tex]a_{t}=9.361[/tex] [tex]m/s^{2}[/tex]
Now calculate radius of curvature to figure out normal force.
[tex]r=((1+(dy/dx)^{2})/(d^{2}y/ d^{2}x )) ^{3/2}[/tex]
Substitute values
[tex]r=94.2[/tex] [tex]m[/tex]
[tex]a_{n} =v_{n} ^{2} /r[/tex]
[tex]v_{n}=4[/tex] [tex]m/s[/tex]
[tex]a_{n}=.169[/tex] [tex]m/s^{2}[/tex]
∑[tex]F_{n}=m*a_{n}[/tex]
[tex]N-,mgcos[/tex]Θ=[tex]m*a_{n}[/tex]
Normal force=[tex]N=310.34[/tex] [tex]N[/tex]
