A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ 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 $8 \frac{m}{s}$ $8 m/s$ 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 · 2017-07-29T04:07:26

$distance = 1.87$ $"distance" = 1.87$ $m$ $"m "$ (if $θ = \frac{π}{3}$ $theta = pi/3$ )

Explanation:

Since there were differing answers here, I thought I'd settle it:)

I'd like to point out that there can't realistically be an inclined ramp with angle of inclination $\frac{2 π}{3}$ $(2pi)/3$ ...it must be between $0$ $0$ and $\frac{π}{2}$ $pi/2$ , so I'll choose the closest angle to this, $\frac{π}{3}$ $pi/3$ ...

We're asked to find the distance traveled by the box given its initial speed, coefficient of kinetic friction, and angle of inclination.

![upload.wikimedia.org]()

NOTE: the friction force actually acts down the incline as opposed to the image (because the box is traveling up the ramp).

The magnitude of the kinetic friction force $f_{k}$ $f_k$ is given by

$f_{k} = μ_{k} n$ $f_k = mu_kn$

where

$μ_{k}$ $mu_k$ is the coefficient of kinetic friction ( $\frac{7}{4}$ $7/4$ )
$n$ $n$ is the magnitude of the normal force exerted by the incline plane, equal to $m g cos θ$ $mgcostheta$

We must first find the acceleration of the box, using Newton's second law:

$\sum F = m a$ $sumF = ma$

$a = \frac{\sum F}{m}$ $a = (sumF)/m$

The net horizontal force $\sum F$ $sumF$ is

$\sum F = m g sin θ + f_{k} = m g sin θ + μ_{k} m g cos θ$ $sumF = mgsintheta + f_k = mgsintheta + mu_kmgcostheta$

Therefore, we have

$a = (cancel(m)gsintheta + cancel(m)u_kmgcostheta)/(cancel(m)) = gsintheta + mu_kgcostheta$

Plugging in known values, we have

$a = (9.81color(white)(l)"m/s"^2)sin((pi)/3) + 7/4(9.81color(white)(l)"m/s"^2)cos((pi)/3)$

$= 17.1$ $"m/s"^2$

directed down the incline, so this can also be written as

$a = ul(-17.1$ $"m/s"^2$

Now, we can use the equation

$(v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)$

to find the distance it travels up the ramp before it comes to a stop.

Here,

$v_x = 0$ (instantaneously at rest at maximum height)

$v_(0x) = 8$ $"m/s"$
$a_x = -17.1$ $"m/s"^2$
$Deltax =$ trying to find

Plugging in known values, we have

$0 = (8color(white)(l)"m/s")^2 + 2(-17.1color(white)(l)"m/s"^2)(Deltax)$

$Deltax = color(red)(ul(1.87color(white)(l)"m"$

Answer 2 · 2017-07-29T06:29:20

I have an answer 1.875m.

Explanation:

Here's explanation.👇
my notebook for explanation

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A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ 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 8 m/s8ms8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 7/4 747/4 and an incline of (2 pi )/3 2π3(2 pi )/3 . How far along the ramp will the box go?

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Explanation:

Explanation:

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A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ 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?