A box with an initial speed of $5 \frac{m}{s}$ $5 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{2}$ $7/2$ 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 $5 \frac{m}{s}$ $5 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{2}$ $7/2$ 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 · 2017-05-14T05:39:57

The distance is $= 0.68 m$ $=0.68m$

Explanation:

Taking the direction up and parallel to the plane as positive $↗^{+}$ $↗^+$

The coefficient of kinetic friction is $μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

Then the net force on the object is

$F = - F_{r} - W sin θ$ $F=-F_r-Wsintheta$

$= - F_{r} - m g sin θ$ $=-F_r-mgsintheta$

$= - μ_{k} N - m g sin θ$ $=-mu_kN-mgsintheta$

$= m μ_{k} g cos θ - m g sin θ$ $=mmu_kgcostheta-mgsintheta$

According to Newton's Second Law

$F = m \cdot a$ $F=m*a$

Where $a$ $a$ is the acceleration

So

$m a = - μ_{k} g cos θ - m g sin θ$ $ma=-mu_kgcostheta-mgsintheta$

$a = - g (μ_{k} cos θ + sin θ)$ $a=-g(mu_kcostheta+sintheta)$

$a = - 9.8 \cdot (\frac{7}{2} cos (\frac{5}{12} π) + sin (\frac{5}{12} π))$ $a=-9.8*(7/2cos(5/12pi)+sin(5/12pi))$

$= - 18.34 m s^{- 2}$ $=-18.34ms^-2$

The negative sign indicates a deceleration

We apply the equation of motion

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$u = 5 m s^{- 1}$ $u=5ms^-1$

$v = 0$ $v=0$

$a = - 18.34 m s^{- 2}$ $a=-18.34ms^-2$

$s = \frac{v^{2} - u^{2}}{2 a}$ $s=(v^2-u^2)/(2a)$

$= \frac{0 - 25}{- 2 \cdot 18.34}$ $=(0-25)/(-2*18.34)$

$= 0.68 m$ $=0.68m$

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A box with an initial speed of $5 \frac{m}{s}$ $5 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{2}$ $7/2$ 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 5 m/s5ms5 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 7/2 727/2 and an incline of (5 pi )/12 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 $5 \frac{m}{s}$ $5 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{7}{2}$ $7/2$ and an incline of $\frac{5 π}{12}$ $(5 pi )/12$ . How far along the ramp will the box go?