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 $\frac{3}{4}$ $3/4$ and an incline of $\frac{π}{8}$ $pi /8$ . 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 $\frac{3}{4}$ $3/4$ and an incline of $\frac{π}{8}$ $pi /8$ . How far along the ramp will the box go?

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Answer 1 · 2018-03-04T09:01:24

The distance is $= 0.05 m$ $=0.05m$

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 $μ_{k} = \frac{3}{4}$ $mu_k=3/4$

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

The incline of the ramp is $θ = \frac{1}{8} π$ $theta=1/8pi$

The acceleration is $a = - 9.8 \cdot (\frac{3}{4} cos (\frac{1}{8} π) + sin (\frac{1}{8} π))$ $a=-9.8*(3/4cos(1/8pi)+sin(1/8pi))$

$= - 10.54 m s^{- 2}$ $=-10.54ms^-2$

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 $a = - 10.54 m s^{- 2}$ $a=-10.54ms^-2$

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

$= \frac{0 - 1}{- 2 \cdot 10.54}$ $=(0-1)/(-2*10.54)$

$= 0.05 m$ $=0.05m$

Answer 2 · 2018-03-04T09:04:42

Here,downward component of the weight of the box which tries to pull it down along the plane is $m g sin (\frac{π}{8}) = 0.383 m g$ $mgsin(pi/8)=0.383mg$

And,maximum value of kinetic frictional force that can act is $μ \times N = \frac{3}{4} m g cos (\frac{π}{8}) = 0.693 m g$ $mu×N=3/4 mg cos (pi/8)=0.693mg$

Now, initially,the box has a tendency to go up,so frictional force will act along with the downward component of its weight to stop the motion.

So,net acceleration downwards will be $1.076 g$ $1.076g$

So,if it goes up by $x m$ $xm$ ,we can say, $0^{2} = 1^{2} - 2 \times 1.076 g x$ $0^2=1^2 -2×1.076gx$

Or, $x = 0.05 m$ $x=0.05m$

After that the block will come to momentary rest and try to move down due to its downwards component of weight,but maximum frictional force value is more than that,so it will keep the block at rest at that point.

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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 $\frac{3}{4}$ $3/4$ and an incline of $\frac{π}{8}$ $pi /8$ . 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 $\frac{3}{4}$ $3/4$ and an incline of $\frac{π}{8}$ $pi /8$ . How far along the ramp will the box go?