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 $\frac{5}{4}$ $5/4$ and an incline of $\frac{π}{6}$ $pi /6$ . How far along the ramp will the box go?

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Answer 1 · 2017-04-22T08:21:24

The distance is $= 0.13 m$ $=0.13m$

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Solving in the direction of the plane $↗^{+}$ $↗^+$

$μ_{k} = \frac{F_{k}}{N}$ $mu_k=F_k/N$

$F_{k} = μ_{k} N$ $F_k=mu_kN$

$N = m g cos θ$ $N=mgcostheta$

$F_{k} = μ_{k} m g cos θ$ $F_k=mu_kmgcostheta$

The component of the weight is

$= m g sin θ$ $=mgsintheta$

Applying Newton' second Law

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

Therefore,

$a = - u_{k} g cos θ - g sin θ$ $a=-u_kgcostheta-gsintheta$

$= - \frac{5}{4} \cdot g \cdot cos (\frac{1}{6} π) - g \cdot sin (\frac{1}{6} π)$ $=-5/4*g*cos(1/6pi)-g*sin(1/6pi)$

$= - 15.51 m s^{- 2}$ $=-15.51ms^-2$

The initial velocity is $u = 2 m s^{- 1}$ $u=2ms^-1$

We apply the equation of motion

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

$0 = 2^{2} - 2 \cdot 15.51 s$ $0=2^2-2*15.51s$

$s = \frac{4}{2 \cdot 15.51}$ $s=4/(2*15.51)$

$= 0.13 m$ $=0.13m$

A box with an initial speed of 2ms2 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 545/4 and an incline of π6pi /6 . How far along the ramp will the box go?