A box with an initial speed of $2 \frac{m}{s}$ $2 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{3}{2}$ $3/2$ 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 $2 \frac{m}{s}$ $2 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{3}{2}$ $3/2$ 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 · 2017-08-05T19:07:18

$distance = 0.126$ $"distance" = 0.126$ $m$ $"m"$

Explanation:

We're asked to find how far the box will travel up the ramp, given its initial speed, the coefficient of kinetic friction, and the angle of inclination.

I will solve this problem using only Newton's laws and kinematics (i.e. without using work/energy).

NOTE: Ideally, the angle of inclination would be between $0$ $0$ and $\frac{π}{2}$ $pi/2$ , so I'll choose the corresponding first-quadrant angle of $\frac{π}{3}$ $pi/3$ .

I will also take the positive $x$ $x$ -direction as up the ramp.

When the box reaches its maximum distance, the instantaneous velocity will be $0$ $0$ . We are ultimately going to use the constant-acceleration equation

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

where

$v_x$ is the instantaneous velocity (which is $0$ )
$v_(0x)$ is the initial velocity
$a_x$ is the (constant) acceleration
$Deltax$ is the distance it travels (what we're trying to find)

Since the velocity $v_x = 0$ , we can also write this equation as

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

We also figure that the acceleration will be negative because it slows down and comes to a brief stop at its maximum height, so we then have

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

And we can move it to the other side:

$ul(2a_x(Deltax) = (v_(0x))^2$

Rearranging for the distance traveled $Deltax$ :

$ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2a_x)" ")|)$

We already know the initial velocity, so we need to find the acceleration of the box.

$" "$

Let's use Newton's second law of motion to find the acceleration, which is

$ul(sumF_x = ma_x$

where

$sumF_x$ is the net force acting on the box
$m$ is the mass of the box
$a_x$ is the acceleration of the box (what we're trying to find)

The only forces acting on the box are

the gravitational force (acting down the ramp), equal to $mgsintheta$
the friction force (acting down the ramp), equal to $mu_kn$

And so we have our net force equation:

$sumF_x = mgsintheta + mu_kn$

The normal force $n$ exerted by the incline is equal to

$n = color(purple)(mgcostheta)$

So we can plug this in to the net force equation above:

$ul(sumF_x = mgsintheta + mu_kcolor(purple)(mgcostheta)$

Or

$ul(sumF_x = mg(sintheta + mu_kcostheta)$

Now, we can plug this in for $sumF_x$ in the Newton's second law equation:

$sumF_x = ma_x$

$ma_x = mg(sintheta + mu_kcostheta)$

We can cancel the mass $m$ by dividing both sides by $m$ , leaving us with

$color(green)(ul(a_x = g(sintheta + mu_kcostheta)$

$" "$

Now that we have found an expression for the acceleration, let's plug it into the equation

$Deltax = ((v_(0x))^2)/(2a_x)$

And we get

$color(red)(ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2g(sintheta + mu_kcostheta))" ")|)$

We're given in the problem

$v_(0x) = 2$ $"m/s"$
$theta = (pi)/3$
$mu_k = 3/2$
and the gravitational acceleration $g = 9.81$ $"m/s"$

Plugging these in:

$color(blue)(Deltax) = ((2color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(pi)/3] + 3/2cos[(pi)/3])) = color(blue)(ulbar(|stackrel(" ")(" "0.126color(white)(l)"m"" ")|)$

Answer 2 · 2017-08-05T21:27:44

Here's my own attempt at this problem, using an alternative approach to Nathan's, to see if our answers agree. Like Nathan, I will assume that you mean to have the ramp right-side-up, i.e. an angle of $60^@$ w.r.t. the horizontal...

I also got $"0.126 m"$ .

My approach here is via conservation of energy:

$DeltaE = DeltaK + DeltaU + W_(vecF_k) = 0$

where:

My coordinate axes are almost the usual vertical $y$ and horizontal $x$ . Rightwards is $+x$ , but downwards is $+y$ .

$DeltaK = 1/2 mv_f^2 - 1/2 mv_i^2 = m/2(v_f^2 - v_i^2)$

$DeltaU = mvecgDeltavecy = mgy_f$ , where $y_i = 0$ , $y_f < 0$ , and $g > 0$ . The potential energy becomes more negative due to the sign of $y_f$ .

$W_(vecF_k) = vecF_kd$ is the counteracting work due to kinetic friction (negative w.r.t. the box, the system). In this case, I define

$d = sqrt((Deltax)^2 + (Deltay)^2) - 0 = sqrt(x_f^2 + y_f^2)$ with $x_i = 0$ and $x_f > 0$ . $d$ is then the final position, $< 0$ .

Currently, we know we have a nonzero initial velocity and a zero final velocity when the box comes to a stop:

$=> DeltaK = -m/2v_i^2$

This so far gives:

$-m/2v_i^2 + mgy_f + vecF_kd = 0$

or

$m/2v_i^2 - mgy_f - vecF_kd = 0$

Using the sum of the forces:

$sum_i vecF_(_|_,i) = vecF_N - mvecgcostheta = 0$

So...

$0 = cancel(m)/2v_i^2 - cancel(m)gy_f - mu_kcancel(m)vecgcosthetad$

$=> 1/2 v_i^2 - gy_f - mu_kgcosthetad = 0$

Now, we can rewrite $y_f$ in terms of $d$ , the distance traveled along the ramp:

$y_f = dsintheta < 0$

This gives:

$=> 1/2 v_i^2 - gdsintheta - mu_kgdcostheta = 0$

Solve for $d$ to get:

$d = (1/2 v_i^2)/(gsintheta + mu_kgcostheta)$

$= (v_i^2)/(2g(sintheta + mu_kcostheta))$

(which is also what Nathan found, by the way.)

So, in the end, we get:

$color(blue)(|d|) = (2^2)/(2(9.81)(sin(60^@) + 3/2cos(60^@)))$

$=$ $color(blue)("0.126 m")$

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

2 Answers

Explanation:

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