A box with an initial speed of $9 \frac{m}{s}$ $9 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{4}$ $5/4$ and an incline of $\frac{3 π}{4}$ $(3 pi )/4$ . How far along the ramp will the box go?

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A box with an initial speed of $9 \frac{m}{s}$ $9 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{4}$ $5/4$ and an incline of $\frac{3 π}{4}$ $(3 pi )/4$ . How far along the ramp will the box go?

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Answer 1 · 2017-02-18T05:53:43

The box will go $- 23.36 m$ $-23.36 m$ etres along the ramp.

Explanation:

If a block is moving up the inclined plane then the distance travelled by the block up the plane is
$s = {\frac{u^{2}}{2 g (sin θ + μ_{k} cos θ)}}$ $s={[u^2]/[2g(sintheta+mu_kcostheta)]}$
Assume box as block and ramp as inclined plane and also consider $g = 9.8 m / s^{2}$ $g=9.8m//s^2$ .
Calculating with given values,
$s = ⎧ ⎪ ⎨ ⎪ ⎩ \frac{{[9 m / s]}^{2}}{2 \times (9.8 m / s^{2}) \times [sin (\frac{3 π}{4}) + (\frac{5}{4}) cos (\frac{3 π}{4})]} ⎫ ⎪ ⎬ ⎪ ⎭$ $s={[9m//s]^2/[2xx(9.8m//s^2)xx[sin((3pi)/4)+(5/4)cos((3pi)/4)]]}$
$:.s=-23.36m$ .
Here' $-$ ' sign denotes that the body is travelling up the incline plane.

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A box with an initial speed of $9 \frac{m}{s}$ $9 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{4}$ $5/4$ and an incline of $\frac{3 π}{4}$ $(3 pi )/4$ . 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 $9 \frac{m}{s}$ $9 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{4}$ $5/4$ and an incline of $\frac{3 π}{4}$ $(3 pi )/4$ . How far along the ramp will the box go?