Coefficient of friction?

A block of weight $6.1 N$ slides down a slope inclined at ${\tan}^{-} 1 \left(\frac{11}{60}\right)$ to the horizontal. The coefficient of friction between the block and the slope is $\frac{1}{4}$. The block passes through a point A with speed $2 m \cdot {s}^{-} 1$. Find how far the block moves from A before it comes to rest.

Aug 10, 2017

Here's what I got.

Explanation:

The trick here is to realize that the block is actually slowing down as it's moving on the ramp.

In other words, the block is launched with an unknown initial velocity at the top of the ramp and it is decelerating as it's making its way down the ramp. This implies that the frictional force will overpower the projection of the block's weight along the length of the ramp.

Now, the forces that act on the block on the inclined plane can be represented like this In your case, the net force that is acting on the block, ${F}_{\text{net}}$, is oriented up the ramp because the block is slowing down.

This implies that you have

$\overbrace{{F}_{\text{net")^(color(blue)("up the ramp")) = overbrace(" "f" ")^(color(blue)("up the ramp")) - overbrace(G * sin(theta))^(color(blue)("down the ramp"))" "color(blue)("(*)}}}$

Here

• $f$ is the frictional force
• $G = m \cdot g$ is the weight of the block, with $g$ being the gravitational acceleration, which I'll take as $g = {\text{9.8 m s}}^{- 2}$
• $\theta$ is the angle the ramp makes with the horizontal, in your case equal to ${\tan}^{- 1} \left(\frac{11}{60}\right)$

By definition, the frictional force is equal to

$f = \mu \cdot N$

Here

• $\mu$ is the coefficient of friction between the block and the ramp
• $N$ is the normal force

As you can see in the diagram, you have

$N = G \cdot \cos \left(\theta\right)$

which implies that

$f = \mu \cdot G \cdot \cos \left(\theta\right)$

Plug this into equation $\textcolor{b l u e}{\text{(*)}}$ to get

${F}_{\text{net}} = \mu \cdot G \cdot \cos \left(\theta\right) - G \cdot \sin \left(\theta\right)$

Now, if you take $m$ to be the mass of the block, you can say that

${\overbrace{\textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot a}}^{\textcolor{b l u e}{= \text{F"_ "net}}} = \mu \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot g \cdot \cos \left(\theta\right) - \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \cdot g \cdot \sin \left(\theta\right)$

which gets you

a = g * [mucos(theta) - sin(theta)]" "color(blue)("(* *)")

At this point, focus on finding the values of $\cos \left(\theta\right)$ and of $\sin \left(\theta\right)$.

You know that

${\tan}^{- 1} \left(\frac{11}{60}\right) = \theta$

so you can say that

$\tan \left(\theta\right) = \frac{11}{60}$

Since

$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$

you can say that

$\sin \frac{\theta}{\cos} \left(\theta\right) = \frac{11}{60}$

This is equivalent to

${\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = {\left(\frac{11}{60}\right)}^{2}$

and

${\sin}^{2} \left(\theta\right) \cdot {60}^{2} = {\cos}^{2} \left(\theta\right) \cdot {11}^{2}$

Use the fact that

${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$

to rewrite this as

$\left[1 - {\cos}^{2} \left(\theta\right)\right] \cdot {60}^{2} = {\cos}^{2} \left(\theta\right) \cdot {11}^{2}$

You should end up with

$\left\{\begin{matrix}\cos \left(\theta\right) = \sqrt{{60}^{2} / \left({11}^{2} + {60}^{2}\right)} = \frac{60}{61} \\ \sin \left(\theta\right) = \sqrt{{11}^{2} / \left({11}^{2} + {60}^{2}\right)} = \frac{11}{61}\end{matrix}\right.$

Now, you know that when the block passes through a point $\text{A}$ on the ramp, its velocity is equal to ${\text{2 m s}}^{- 1}$.

If you take into account that the block will stop at a distance $d$ after it passes through point $\text{A}$, i.e. its final velocity will be equal to ${\text{0 m s}}^{- 1}$, you can say that

${0}^{2} = {v}_{\text{A}}^{2} - 2 \cdot a \cdot d$

Rearrange to solve for $d$

${v}_{\text{A"^2 = 2 * a * d implies d = v_"A}}^{2} / \left(2 \cdot a\right)$

Finally, use equation $\textcolor{b l u e}{\text{(* *)}}$ to find

d = (2^2 "m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-2)))))/( 2 * 9.8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2)))) * (1/4 * 60/61 - 11/61))

$d = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{3.1 m}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the velocity of the block as it passes through point $\text{A}$.