A box with an initial speed of $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $5/6$ and an incline of $pi /3$ . How far along the ramp will the box go?

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A box with an initial speed of $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $5/6$ and an incline of $pi /3$ . How far along the ramp will the box go?

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Answer 1 · 2016-11-10T20:23:28

$s=v^2/(2g(sin alpha+ k cos alpha))$

$s=0.64m$

Explanation:

$E_k=E_p+W_T$
$mv^2/2=mgh+F_ts$
$F_t=F_n=mg*cos alpha$
$h/s=sin alpha$
$mv^2/2=mgs *sin alpha+kmgs cos alpha$
$s=v^2/(2g(sin alpha+ k cos alpha))$
$k=5/6$
$alpha=pi/3$
$g=9.81m/s^2$
$s=0.64m$

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A box with an initial speed of $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $5/6$ and an incline of $pi /3$ . How far along the ramp will the box go?

1 Answer

Explanation:

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A box with an initial speed of 4 m/s4 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/6 5/6 and an incline of pi /3 pi /3 . How far along the ramp will the box go?

1 Answer

Explanation:

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A box with an initial speed of $4 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $5/6$ and an incline of $pi /3$ . How far along the ramp will the box go?