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 $\frac{1}{2}$ $1/2$ 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 · 2017-04-12T06:03:27

The distance is $= 1.14 m$ $=1.14m$

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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,

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

$= - \frac{1}{2} \cdot g \cdot cos (\frac{3}{8} π) - g \cdot sin (\frac{3}{8} π)$ $=-1/2*g*cos(3/8pi)-g*sin(3/8pi)$

$= - 10.93 m s^{- 2}$ $=-10.93ms^-2$

The initial velocity is $u = 5 m s^{- 1}$ $u=5ms^-1$

We apply the equation of motion

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

$0 = 5^{2} - 2 \cdot 10.93 s$ $0=5^2-2*10.93s$

$s = \frac{25}{2 \cdot 10.93}$ $s=25/(2*10.93)$

$= 1.14 m$ $=1.14m$

A box with an initial speed of 5ms5 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 121/2 and an incline of 3π8(3 pi )/8 . How far along the ramp will the box go?