Respuesta :
Answer:
(a). The value of distance is 5.043 m
(b). The coefficient of friction is 0.302.
(c). The distance is 10.33 m
Explanation:
Given that,
Speed = 4.50 m/s
Minimum friction = 0.1
Maximum friction = 0.6
Distance = 12.5 m
We know that,
The newton's second law
[tex]F_{net}=-\mu N[/tex]
[tex]F_{net}=-\mu mg[/tex]
We know that,
The frictional coefficient is directly proportional to the distance.
We need to calculate the value of coefficient
Using general equation of frictional coefficient
[tex]\mu(x)=Ax+B[/tex]....(I)
At x= 0,
[tex]\mu(0)=0+B[/tex]
Put the value into the formula
[tex]B=0.1[/tex]
Now for value of A
Put the value in equation (I)
[tex] 0.6=A12.5+0.1[/tex]
[tex]A=\dfrac{1}{25}[/tex]
Put the value in equation (I)
[tex]\mu(x)=\dfrac{x}{25}+0.1[/tex]......(II)
We need to calculate the work done
Using work energy theorem
[tex]W=\int_{0}^{x}{F_{net} dx}[/tex]
[tex]W=\int_{0}^{x}{-\mu_{x}mgdx}[/tex]
[tex]W=-mg\int_{0}^{x}{(\dfrac{x}{25}+0.1)dx}[/tex]
[tex]W=-mg(\dfrac{x^2}{50}+0.1x)[/tex]....(III)
(a). We need to calculate the distance
Using difference of kinetic energy
[tex]W=\dfrac{1}{2}m(v_{f}^2-v_{0}^2)[/tex]
[tex]-mg(\dfrac{x^2}{50}+0.1x)=\dfrac{1}{2}m(0-(4.50)^2)[/tex]
[tex]\dfrac{x^2}{50}+0.1x=1.013[/tex]
[tex]x^2+5x=50.65[/tex]
[tex]x^2+5x-50.65=0[/tex]
[tex]x=5.043\ m[/tex]
(b). We need to calculate the coefficient of friction at the stopping point
Using equation (III)
[tex]\mu(5.043)=\dfrac{5.043}{25}+0.1[/tex]
[tex]\mu(5.043)=0.302[/tex]
(c). We need to calculate the distance
Using formula work done
[tex]W=-fx_{2}[/tex]
[tex]W=-\mu mgx_{2}[/tex]
[tex]-\dfrac{1}{2}mv_{0}^2=-\mu mgx_{2}[/tex]
[tex]x_{2}=\dfrac{v_{0}^2}{2\mu g}[/tex]
Put the value into the formula
[tex]x_{2}=\dfrac{(4.5)^2}{2\times0.1\times9.8}[/tex]
[tex]x_{2}=10.33\ m[/tex]
Hence, (a). The value of distance is 5.043 m
(b). The coefficient of friction is 0.302.
(c). The distance is 10.33 m
