Question

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 1

Answer:

0.4998 J

Explanation:

Mass if hockey puck=m=0.17 kg

$\mu(x)=0.1+0.05x$

Force acting on hockey puck=$F=\mu mg$

Where $g=9.8m/s^2$

Work done =Fdx

x=0 to x=2 m

Using the formula

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

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

Using the formula

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Work done=W=$1.666(0.1(2)+\frac{0.05}{2}(2)^2)=0.4998 J$