A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{6}$ $5/6$ and an incline of $\frac{π}{4}$ $pi /4$ . How far along the ramp will the box go?

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A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{6}$ $5/6$ and an incline of $\frac{π}{4}$ $pi /4$ . How far along the ramp will the box go?

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Answer 1 · 2018-03-04T10:04:25

The distance is $= 0.04 m$ $=0.04m$

Explanation:

Resolving in the direction up and parallel to the plane as positive $↗^{+}$ $↗^+$

The coefficient of kinetic friction is $μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

Then the net force on the object is

$F = - F_{r} - W sin θ$ $F=-F_r-Wsintheta$

$= - F_{r} - m g sin θ$ $=-F_r-mgsintheta$

$= - μ_{k} N - m g sin θ$ $=-mu_kN-mgsintheta$

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

According to Newton's Second Law of Motion

$F = m \cdot a$ $F=m*a$

Where $a$ $a$ is the acceleration of the box

So

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

$a = - g (μ_{k} cos θ + sin θ)$ $a=-g(mu_kcostheta+sintheta)$

The coefficient of kinetic friction is $μ_{k} = \frac{5}{6}$ $mu_k=5/6$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The incline of the ramp is $θ = \frac{1}{4} π$ $theta=1/4pi$

The acceleration is $a = - 9.8 \cdot (\frac{5}{6} cos (\frac{1}{4} π) + sin (\frac{1}{4} π))$ $a=-9.8*(5/6cos(1/4pi)+sin(1/4pi))$

$= - 12.7 m s^{- 2}$ $=-12.7ms^-2$

The negative sign indicates a deceleration

Apply the equation of motion

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

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

The final velocity is $v = 0$ $v=0$

The acceleration is $a = - 12.7 m s^{- 2}$ $a=-12.7ms^-2$

The distance is $s = \frac{v^{2} - u^{2}}{2 a}$ $s=(v^2-u^2)/(2a)$

$= \frac{0 - 1}{- 2 \cdot 12.7}$ $=(0-1)/(-2*12.7)$

$= 0.04 m$ $=0.04m$

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A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{6}$ $5/6$ and an incline of $\frac{π}{4}$ $pi /4$ . How far along the ramp will the box go?

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Explanation:

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A box with an initial speed of $1 \frac{m}{s}$ $1 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{5}{6}$ $5/6$ and an incline of $\frac{π}{4}$ $pi /4$ . How far along the ramp will the box go?