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

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Answer 1 · 2017-04-01T17:02:02

The distance is $= 1.98 m$ $=1.98m$

The coefficient of kinetic friction is

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

Let the mass of the object be $= m k g$ $=m kg$

The incline is $= θ$ $=theta$

Therefore,

$F_{r} = μ_{k} \cdot N = μ_{k} \cdot m g cos θ$ $F_r=mu_k*N=mu_k*mg costheta$

The net force on the box is

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

According to Newtons' second Law

$F = m a$ $F=ma$

Where the acceleration is $= a h$ $=ah$

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

$a = - \frac{2}{5} g cos (\frac{π}{12}) - g sin (\frac{π}{12})$ $a=-2/5gcos(pi/12)-gsin(pi/12)$

$a = - 6.32 m s^{- 2}$ $a=-6.32ms^-2$

We apply the equation of motion

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

$v = 0$ $v=0$

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

So,

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

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

$= \frac{25}{2 \cdot 6.32} = 1.98 m$ $=25/(2*6.32)=1.98m$

A box with an initial speed of 55 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 252/5 and an incline of π12pi /12 . How far along the ramp will the box go?