A particle moving on a horizontal dry surface is subject to a frictional force in the direction opposite to its velocity. There is no frictional force if the object is stationary. The magnitude of the frictional force is assumed to be equal to a coefficient of friction times the weight

Question

A particle moving on a horizontal dry surface is subject to a frictional force in the direction opposite to its velocity. There is no frictional force if the object is stationary. The magnitude of the frictional force is assumed to be equal to a coefficient of friction times the weight (gravitational force) of the object. Find an expression for the position and velocity of an object of mass 1.000 kg if it moves in the x direction with an initial velocity [latex]v_{x}(0)=10.0ms^{−1}[/latex] and an initial position of x=0 and if the coefficient of friction is equal to 0.800.

Accepted Answer

So long as [latex]v_{x}[/latex] is nonzero, the frictional force is

[latex]F_{x}=−(0.800)(1.000 kg) \bigl(9.80ms^{−2}\bigr)=−7.84 kgms^{−2}=−7.84N[/latex]

Newton’s second law gives the equation of motion:

[latex]F_{x}=m\frac{d^{2}x}{dt^{2}} =m\frac{dv_{x}}{dt} [/latex]

so that

[latex]\frac{dv_{x}}{dt} =-\frac{F_{x}}{m} =-\frac{7.84 kgms^{-2}}{1.00 kg} =−7.84 ms^{-2}[/latex]

We integrate this equation from t=0 to t=t′:

[latex]\int_{0}^{t^{′}}{\frac{dv_{x}}{dt} dt} =-\bigl(7.84ms^{−2}\bigr) \int_{0}^{t^{′}}{dt}[/latex]

[latex]v_{x}(t^{′})−v_{x}(0)=−(7.84ms^{−2})t^{′}[/latex]

[latex]vx(t^{′})=10.0ms^{−1}−(7.84ms^{−2})t^{′}[/latex]

This solution applies only until [latex]v_{x}[/latex] vanishes, at which time the force stops and the particle remains stationary. This occurs when

[latex]t^{′}=\frac{10.0ms^{−1}}{7.84ms^{−2}} =1.28 s[/latex]

The position is given by another integration:

[latex]x(t^{′})-x(0)=\int_{0}^{t^{′}}{\frac{dx}{dt} dt} =\int_{0}^{t^{′}}{v_{x}dt} =\int_{0}^{t^{′}}{\left[10.0ms^{−1} \bigl(7.84ms^{−2}\bigr) \right] } dt[/latex]

In this equation we have used the expression for the velocity in which t′ was replaced by t. The integration yields

[latex]x(t^{′})=\bigl(10.0ms^{−1}\bigr) t^{′}-\frac{1}{2} \bigl(7.84ms^{−2}\bigr)t^{′2}[/latex]

This solution applies only for values of t′ no larger than 1.28 s. The position and velocity are shown in Figure 9.1.

Question 9.1: A particle moving on a horizontal dry surface is subject to ...

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