A box with an initial speed of $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $5/2$ and an incline of $pi /4$ . How far along the ramp will the box go?

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Answer 1 · 2017-04-09T07:11:23

The distance is $=0.021m$

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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,

$a=-u_kgcostheta-gsintheta$

$=-5/2*g*cos(1/4pi)-g*sin(1/4pi)$

$=-24.3ms^-2$

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

We apply the equation of motion

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

$0=1-2*24.3*s$

$s=1/(2*24.3)$

$=0.021m$

A box with an initial speed of 1 m/s1 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/2 5/2 and an incline of pi /4 pi /4 . How far along the ramp will the box go?