Answer:
a)[tex]W_{f}=307.87 J[/tex]
b)[tex]W_{f}=615.752 J[/tex]
c) Therefore it is a nonconservative force
Explanation:
Distance traveled in one complete circular trip against friction
[tex]d=2\pi *r[/tex]
The force of friction is
[tex]F_{f}=u*F_{N}[/tex]
[tex]F_{N}=w*g[/tex]
The work of friction is the force of friction in the distance so
[tex]W_{f}=-F_{f}*d[/tex]
a).
[tex]F_{N}=10kg*9.8\frac{m}{s^2}[/tex]
[tex]F_{N}=98N[/tex]
[tex]F_{f}=0.250*98N[/tex]
[tex]F_{f}=24.5N[/tex]
[tex]W_{f}=-24.5N*2\pi*2m[/tex]
[tex]W_{f}=307.87 J[/tex]
b).
[tex]W_{f}=-24.5N*2\pi*4m[/tex]
[tex]W_{f}=615.752 J[/tex]
c).
Since the circle returns to its initial position in both cases and the work done by the friction force different in both cases, so the friction force does depend on the path;
