AP Physics 1 2.7.A: Kinetic Friction

(c) 4 points

For using kinematic data to obtain acceleration, a, or showing that a=0

Alternatively the point can be earned using the Work-Energy solution with kinetic

energy $K=(1 / 2) m v^{2}$

For relating frictional force and coefficient of kinetic friction as $F_{f}=\mu_{K} F_{N}$

For a correct representation of the normal force, correctly referring back to a

measurement of the fan-block system

$F_{N}=m g$ if $\theta=0$ or $F_{N}=m g+F \sin \theta$ if $\theta \neq 0$

For using Newton's second law, consistently with normal force above, for the

appropriate forces or force components in the direction of the fan's motion that

would lead to the determination of the coefficient of kinetic friction

$F_{\text {net }}=m a$

$F-F_{f}=m a$

$\mu_{k}=\frac{F-m a}{m g}$ if $\theta=0$ or $\mu_{k}=\frac{F \cos \theta-m a}{m g+F \sin \theta}$ if $\theta \neq 0$

Alternatively the point can be earned using the Work-Energy solution that would lead

to the determination of the coefficient of kinetic friction.

Example 3 (continued from above): $\theta=0$ Determine the slope of each v vs. t graph,

to get the acceleration for each trial, and average the accelerations. Call this average

acceleration a. The kinetic frictional force exerted by track on block is

$F_{f}=\mu F_{n}=\mu m g$, where m is the mass of the fan-block as found in part (a), since

the acceleration is purely horizontal and hence $F_{n}$ and m g must balance. So, by

Newton's second law, letting F denote the force exerted by the air on the fan found

in part (a), we get

Solve for $\mu$ to get $\mu=\frac{F-m a}{m g}$

Example 4 (continued from above): $\theta=0$ Let d denote distance from block to second

photogate and t denote the corresponding time. Since $v_{0}=0, d=\frac{1}{2} a t^{2}$ and

hence $a=\frac{2 d}{t^{2}}$. For each trial, calculate that acceleration, and take the average,

which becomes our best estimate of a. [Rest of solution is as in example 3]