A box with an initial speed of $6 \frac{m}{s}$ $6 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{5}$ $2/5$ and an incline of $\frac{5 π}{12}$ $( 5 pi )/12$ . How far along the ramp will the box go?

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A box with an initial speed of $6 \frac{m}{s}$ $6 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{5}$ $2/5$ and an incline of $\frac{5 π}{12}$ $( 5 pi )/12$ . How far along the ramp will the box go?

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Answer 1 · 2017-07-12T06:41:33

Ans: 1.9 m

Explanation:

Given:
u = 6m/s.
$μ$ $mu$ =2/5 = 0.6
$θ = \frac{5 π}{12} = 75 d e g r e e s$ $theta =( 5 pi)/12= 75 degrees$

For an inclined plane,

$a = g (sin θ - μ cos θ)$ $a = g(sin theta - mu cos theta)$
$a = 9.8 (sin 75 - 0.4 \cdot cos 75)$ $a= 9.8 ( sin 75- 0.4 * cos75)$
$a = 8.45 \frac{m}{s^{2}}$ $a= 8.45 m/(s^2)$

Again, $v = u - (a sin θ) t$ $v= u -( asin theta)t$
$\frac{u}{a sin θ} = t$ $u/(asintheta) = t$
$t = 0.634 s$ $t= 0.634 s$

Thus,
$s = u t - \frac{1}{2} g sin θ \cdot t^{2}$ $s= ut - 1/2 gsintheta * t^2$

Therefore, $s = 1.9 m$ $s= 1.9 m$

Thank you!!

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A box with an initial speed of $6 \frac{m}{s}$ $6 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{5}$ $2/5$ and an incline of $\frac{5 π}{12}$ $( 5 pi )/12$ . 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 6ms6 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 252/5 and an incline of 5π12( 5 pi )/12 . How far along the ramp will the box go?

1 Answer

Explanation:

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A box with an initial speed of $6 \frac{m}{s}$ $6 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{5}$ $2/5$ and an incline of $\frac{5 π}{12}$ $( 5 pi )/12$ . How far along the ramp will the box go?