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A hockey puck of mass 0.17 kg is shot across a rough floor with the roughness different at different places, which can be described by a position-dependent coefficient of kinetic friction. For a puck moving along the x-axis, the coefficient of kinetic friction is the following function of x, where x is in μ(x)=0.1+0.05x. Find the work done by the kinetic frictional force on the hockey puck when it has moved from x=0 to x=2 m.

Answer :

lublana

Answer:

0.4998 J

Explanation:

Mass if hockey puck=m=0.17 kg

[tex]\mu(x)=0.1+0.05x[/tex]

Force acting on hockey puck=[tex]F=\mu mg[/tex]

Where [tex]g=9.8m/s^2[/tex]

Work done =Fdx

x=0 to x=2 m

Using the formula

Work done=[tex]W=\int_{0}^{2}(0.17)(9.8)(0.1+0.05x)dx[/tex]

Work done=[tex]W=1.666[0.1x+\frac{0.05}{2}x^2]^{2}_{0}[/tex]

Using the formula

[tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]

Work done=W=[tex]1.666(0.1(2)+\frac{0.05}{2}(2)^2)=0.4998 J[/tex]

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