Question

y - components of newtons second law then give two equations and two unknowns for the coefficient of kinetic friction, $mu_{k}$, and the normal force $n$.

solution

solve the time - independent kinematic equation for the acceleration $a$. $v^{2}=v_{0}^{2}+2adelta x$ $a = \frac{v^{2}-v_{0}^{2}}{2delta x}$

substitute $v = 0$, $v_{0}=20.0 m/s$, and $delta x = 1.20\times10^{2} m$. note the negative sign in the answer: $vec{a}$ is opposite $vec{v}$. $a=\frac{0-(20.0 m/s)^{2}}{2(1.20\times10^{2} m)}=-1.67 m/s^{2}$

find the normal force from the $y$-component of the second law. $sum f_{y}=n - f_{g}=n - mg = 0$ $n = mg$

obtain an expression for the force of kinetic friction, and substitute it into the $x$-component of the second law. $f_{k}=mu_{k}n=mu_{k}mg$ $ma=sum f_{x}=-f_{k}=-mu_{k}mg$

solve for $mu_{k}$ and substitute values. $mu_{k}=-\frac{a}{g}=\frac{1.67 m/s^{2}}{9.80 m/s^{2}} = 0.170$

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remarks notice how the problem breaks down into three parts: kinematics, newtons second law in the $y$-direction, and then newtons law in the $x$-direction.

question if the puck were struck in the same way by an astronaut on a patch of ice on mars, where the acceleration of gravity is $0.35 g$, so that the puck left the hockey stick with the same speed, the distance it travels would be times its distance of travel on earth. (assume that $mu$ remains the same as in the preceding problem.)

Explanation:

Step1: Recall the kinematic - acceleration formula

$a = \frac{v^{2}-v_{0}^{2}}{2\Delta x}$
On Earth, we had $a_{E}=\frac{0-(20.0\ m/s)^{2}}{2(1.20\times 10^{2}\ m)}=- 1.67\ m/s^{2}$ and $\mu_{k}=-\frac{a_{E}}{g_{E}}$, where $g_{E} = 9.80\ m/s^{2}$.
On Mars, $g_{M}=0.35g_{E}$, and since $\mu_{k}$ is the same, $a_{M}=-\mu_{k}g_{M}$.

Step2: Express acceleration on Mars in terms of Earth - acceleration

Since $\mu_{k}=-\frac{a_{E}}{g_{E}}=-\frac{a_{M}}{g_{M}}$, then $a_{M}=\frac{g_{M}}{g_{E}}a_{E}=0.35a_{E}$.

Step3: Use the kinematic formula for distance

From $v^{2}=v_{0}^{2}+2a\Delta x$, when $v = 0$, $\Delta x=-\frac{v_{0}^{2}}{2a}$.
Let $\Delta x_{E}=-\frac{v_{0}^{2}}{2a_{E}}$ and $\Delta x_{M}=-\frac{v_{0}^{2}}{2a_{M}}$.
Then $\frac{\Delta x_{M}}{\Delta x_{E}}=\frac{a_{E}}{a_{M}}$.
Substituting $a_{M}=0.35a_{E}$ into the above - formula, we get $\frac{\Delta x_{M}}{\Delta x_{E}}=\frac{1}{0.35}\approx2.86$.

Answer:

$2.86$