A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{3 π}{8}$ $(3 pi )/8$ . How far along the ramp will the box go?

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A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{3 π}{8}$ $(3 pi )/8$ . How far along the ramp will the box go?

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Answer 1 · 2018-03-02T01:20:32

I would take the sum of the parallel and perpendicular forces. The perpendicular forces ( ${\to F}_{N}$ $vecF_N$ , ${\to F}_{g}$ $vecF_g$ ) cancel out.

As the box moves forward, the kinetic friction force resists it, and the parallel component of the gravitational force resists it too. That means both those forces are negatively-signed with respect to forward motion.

So:

$\sum {\to F}_{∣ ∣} = - {\to F}_{k} - {\to F}_{g, ∣ ∣}$ $sum vecF_(||) = -vecF_k - vecF_(g,||)$

$= - μ_{k} {\to F}_{N} - m \to g sin θ = m {\to a}_{∣ ∣}$ $= -mu_kvecF_N - mvecgsintheta = mveca_(||)$

where we have put the signs in the equation, $μ_{k} = \frac{2}{3}$ $mu_k = 2/3$ , and $\to g > 0$ $vecg > 0$ .

$\sum {\to F}_{⊥} = {\to F}_{N} - {\to F}_{g, ⊥} = {\to F}_{N} - m \to g cos θ = 0$ $sum vecF_(_|_) = vecF_N - vecF_(g,_|_) = vecF_N - mvecgcostheta = 0$

As a result,

${\to F}_{N} = m \to g cos θ$ $vecF_N = mvecgcostheta$

and

$-mu_kcancel(m)vecgcostheta - cancel(m)vecgsintheta = cancel(m)veca_(||)$

Therefore:

$veca_(||) = -(mu_k vecg cos theta + vecg sin theta)$

$= -(2/3 cdot "9.81 m/s"^2 cdot cos((3pi)/8) + "9.81 m/s"^2 cdot sin((3pi)/8))$

$= -"11.57 m/s"^2$

This says that logically, the box will slow down to zero velocity. Since the ramp is straight, we can assume this is the average acceleration:

$veca_(||) -= (Deltavecv_(||))/(Deltax)$

Letting the bottom of the ramp be the initial position $x_i = "0 m"$ and the final velocity being $"0 m/s"$ ,

$veca_(||) = (0 - vecv_i)/(x_f - 0)$

As a result,

$color(blue)(x_f) = -(vecv_i)/(veca_(||)) = -("1 m/s")/(-"11.57 m/s"^2)$

$=$ $color(blue)"0.086 m"$ up the ramp.

That's about $3.4$ inches.

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A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{3 π}{8}$ $(3 pi )/8$ . How far along the ramp will the box go?

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A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{3 π}{8}$ $(3 pi )/8$ . How far along the ramp will the box go?