A box with an initial speed of $4 \frac{m}{s}$ $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{4}$ $7/4$ and an incline of $\frac{2 π}{3}$ $(2 pi )/3$ . How far along the ramp will the box go?

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A box with an initial speed of $4 \frac{m}{s}$ $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{4}$ $7/4$ and an incline of $\frac{2 π}{3}$ $(2 pi )/3$ . How far along the ramp will the box go?

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Answer 1 · 2016-06-28T19:07:06

The box will go about 0.469 meters up the ramp.

Explanation:

The first thing to note is that $\frac{2 π}{3}$ $(2pi)/3$ is actually an angle past vertical. This means we are not going left-to-right but rather right-to-left. What this amounts to is that all of our values are going to be negative.

A simple way to solve this is to simply change $\frac{2 π}{3}$ $(2pi)/3$ to $\frac{π}{3}$ $pi/3$ . This is equivalent to a ramp with the exact same incline, except in the left-to-right direction, and will save us from having to worry about unnecessary negative values while solving.

Next we need to determine the forces and accelerations present in our system. This is easiest drawn out, but since this is a simple problem, it can be explained as well.

$F = m a = m g sin (θ) + μ m g cos (θ)$ $F = ma = mgsin(theta) + mumgcos(theta)$

Note that the $m g sin (θ)$ $mgsin(theta)$ is the contribution from the force of gravity, and the $μ m g cos (θ)$ $mumgcos(theta)$ is the contribution due to the frictional force.

Obviously masses cancel giving our acceleration:
$a = g sin (θ) + μ g cos (θ)$ $a = gsin(theta) + mugcos(theta)$

Plugging in the given values, we can determine the actual value of the acceleration immediately.
$a = 9.8 \cdot (sin (\frac{π}{3}) + μ cos (\frac{π}{3})) = 9.8 \cdot (\frac{\sqrt{3}}{2} + \frac{7}{4} \cdot \frac{1}{2})$ $a = 9.8*(sin(pi/3) + mucos(pi/3)) = 9.8*(sqrt(3)/2 + 7/4*1/2)$

Which is about $a = 17.062$ $a = 17.062$

Now we can solve for how long the box will take to stop using:
$v - a t = 0$ $v - at = 0$

Plugging in values gives:
$4 - 17.062 \cdot t = 0$ $4 - 17.062*t = 0$
$17.062 \cdot t = 4$ $17.062*t = 4$
$t = 0.234$ $t = 0.234$

Knowing the time, we can now solve for the actual answer using the full equation of motion:
$d = v t - \frac{1}{2} a t^{2}$ $d = vt - 1/2at^2$

At this point we need only plug in our values to find the answer:
$d = 4 \cdot 0.234 - \frac{1}{2} \cdot 17.062 \cdot {0.234}^{2}$ $d = 4*0.234 - 1/2*17.062*0.234^2$
$d = 0.936 - 0.467$ $d = 0.936 - 0.467$
$d = 0.469$ $d = 0.469$

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A box with an initial speed of $4 \frac{m}{s}$ $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{4}$ $7/4$ and an incline of $\frac{2 π}{3}$ $(2 pi )/3$ . How far along the ramp will the box go?

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Explanation:

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A box with an initial speed of $4 \frac{m}{s}$ $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{4}$ $7/4$ and an incline of $\frac{2 π}{3}$ $(2 pi )/3$ . How far along the ramp will the box go?