A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{3}{4}$ $3/4$ and an incline of $\frac{5 π}{12}$ $(5 pi )/12$ . How far along the ramp will the box go?

Question

1380 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-06T12:43:29

The distance is $= 2.81 m$ $=2.81m$

enter image source here

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,

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

$= - \frac{3}{4} \cdot g \cdot cos (\frac{5}{12} π) - g \cdot sin (\frac{5}{12} π)$ $=-3/4*g*cos(5/12pi)-g*sin(5/12pi)$

$= - 11.4 m s^{- 2}$ $=-11.4ms^-2$

The initial velocity is $u = 8 m s^{- 1}$ $u=8ms^-1$

We apply the equation of motion

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

$0 = 64 - 2 \cdot 11.4 \cdot s$ $0=64-2*11.4*s$

$s = \frac{64}{2 \cdot 11.4}$ $s=64/(2*11.4)$

$= 2.81 m$ $=2.81m$

A box with an initial speed of 8ms8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 343/4 and an incline of 5π12(5 pi )/12 . How far along the ramp will the box go?