A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{π}{6}$ $pi /6$ . How far along the ramp will the box go?

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A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{π}{6}$ $pi /6$ . How far along the ramp will the box go?

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Answer 1 · 2017-07-20T18:35:58

The distance is $= 3.03 m$ $=3.03m$

Explanation:

Taking 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

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

Where $a$ $a$ is the acceleration

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

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

The coefficient of kinetic friction is $μ_{k} = \frac{2}{3}$ $mu_k=2/3$

The incline of the ramp is $θ = \frac{1}{6} π$ $theta=1/6pi$

$a = - 9.8 \cdot (\frac{2}{3} cos (\frac{1}{6} π) + sin (\frac{1}{6} π))$ $a=-9.8*(2/3cos(1/6pi)+sin(1/6pi))$

$= - 1056 m s^{- 2}$ $=-1056ms^-2$

The negative sign indicates a deceleration

We apply the equation of motion

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

$u = 8 m s^{- 1}$ $u=8ms^-1$

$v = 0$ $v=0$

$a = - 11.5 m s^{- 2}$ $a=-11.5ms^-2$

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

$= \frac{0 - 64}{- 2 \cdot 10.56}$ $=(0-64)/(-2*10.56)$

$= 3.03 m$ $=3.03m$

Answer 2 · 2017-07-21T14:09:42

Distance traveled up the ramp = 3.03087 m (after rounding off)

Explanation:

We can apply energy conservation rule here.

Initial K.E. of the box = Final P.E. of the box + energy lost due to friction

This is because initial P.E. of the box is zero as height is zero.
Also final K.E. is zero because box is finally at rest.

Suppose 'm' is the mass of box, 'u' is the initial velocity ( $8 \frac{m}{s}$ $8 m/s$ ), 'h' is the height through which it raises, 's' is the distance traveled along the ramp which is to be calculated.

Initial K.E. = $\frac{1}{2} m \cdot u^{2}$ $1/2m*u^2$
Final P.E. = mgh
Energy lost due to friction = work done to overcome frictional force
= $F_{k} \cdot s$ $F_k*s$
= $μ_{k} m g s cos (\frac{π}{6})$ $mu_kmgs cos(pi/6)$

Put it together to get

$\frac{1}{2} m \cdot u^{2} = m g h + μ_{k} m g s cos (\frac{π}{6})$ $1/2m*u^2= mgh+mu_kmgscos(pi/6)$ ...............(1)

Cancel the common mass from both sides.
From the right angled triangle (that I am not inserting here) we get
$h = s \cdot sin (\frac{π}{6})$ $h=s*sin(pi/6)$

Simplifying the equation (1) we get,
$s = \frac{\sqrt{3} \cdot u^{2}}{g (2 + \sqrt{3})}$ $s=(sqrt(3)*u^2)/(g(2+sqrt(3)))$

put u = 8 $\frac{m}{s}$ $m/s$ , g = 9.8 $\frac{m}{s^{2}}$ $m/s^2$ , so

$s = 3.03087 m$ $s=3.03087m$

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A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{π}{6}$ $pi /6$ . How far along the ramp will the box go?

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Explanation:

Explanation:

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A box with an initial speed of 8ms8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 232/3 and an incline of π6pi /6 . How far along the ramp will the box go?

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A box with an initial speed of $8 \frac{m}{s}$ $8 m/s$ is moving up a ramp. The ramp has a kinetic friction coefficient of $\frac{2}{3}$ $2/3$ and an incline of $\frac{π}{6}$ $pi /6$ . How far along the ramp will the box go?