A box with an initial speed of $5 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $1/2$ and an incline of $(3 pi )/8$ . How far along the ramp will the box go?

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Answer 1 · 2017-04-12T06:03:27

The distance is $=1.14m$

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

$mu_k=F_k/N$

$F_k=mu_kN$

$N=mgcostheta$

$F_k=mu_kmgcostheta$

The component of the weight is

$=mgsintheta$

Applying Newton' second Law

$-mu_kmgcostheta-mgsintheta=ma$

Therefore,

$a=-u_kgcostheta-gsintheta$

$=-1/2*g*cos(3/8pi)-g*sin(3/8pi)$

$=-10.93ms^-2$

The initial velocity is $u=5ms^-1$

We apply the equation of motion

$v^2=u^2+2as$

$0=5^2-2*10.93s$

$s=25/(2*10.93)$

$=1.14m$

A box with an initial speed of 5 m/s5 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 1/2 1/2 and an incline of (3 pi )/8 (3 pi )/8 . How far along the ramp will the box go?