Find distance up the incline?

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Find distance up the incline?

An 8 kg block is shot up an incline of $30^{\circ}$ $30^@$ with an initial speed of 2.8 m/s. How far up the incline will the block travel if the coefficient of friction between it and the incline is 0.18?
Hint: Express the height above the ground that the block is above the incline in terms of the distance up the incline (hypotenuse).

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Answer 1 · 2017-06-28T00:03:04

I am not going to take the hint, because I am going to rotate the entire system so that the forces are described either normal to the inclined plane or parallel to the inclined plane.

Summing the forces normal to the plane

$- F_{normal} + m \to g cos (30^{\circ}) = 0$ $-F_"normal" + mvecgcos(30^@) = 0$

where $\to g$ $vecg$ is $- 9.8 {m/s}^{2}$ $-9.8" m/s"^2$

$F_{normal} = m \to g cos (30^{\circ})$ $F_"normal" = mvecgcos(30^@)$

Summing the forces parallel to the incline:

$- {\to F}_{friction} + m \to g sin (30^{\circ}) = m \to a [1]$ $-vecF_"friction" + mvecgsin(30^@) = mveca" [1]"$

We know that ${\to F}_{friction} = μ F_{normal}$ $vecF_"friction" = muF_"normal"$

$- {\to F}_{friction} = 0.18 m \to g cos (30^{\circ}) [2]$ $-vecF_"friction" = 0.18mvecgcos(30^@)" [2]"$

Substituting equation [2] into equation [1]:

$0.18 m \to g cos (30^{\circ}) + m \to g sin (30^{\circ}) = m \to a$ $0.18mvecgcos(30^@) + mvecgsin(30^@) = mveca$

$\to a = 0.5 \to g + 0.18 \frac{\sqrt{3}}{2} \to g$ $veca = 0.5vecg+0.18sqrt3/2vecg$

$\to a = (0.5 + 0.18 \frac{\sqrt{3}}{2}) \to g$ $veca = (0.5+0.18sqrt3/2)vecg$

$\to a = (0.5 + 0.18 \frac{\sqrt{3}}{2}) - 9.8 {m/s}^{2}$ $veca = (0.5+0.18sqrt3/2)-9.8" m/s"^2$

$\to a \approx - 6.43 {m/s}^{2}$ $veca ~~ -6.43" m/s"^2$

Please notice that $\to a$ $veca$ is a negative acceleration parallel to the plane, therefore, can use the equation $V_{f}^{2} - V_{i}^{2} = 2 a s$ $V_f^2 - V_i^2 = 2as$

Where $V_{f} = 0$ $V_f= 0$ , $V_{i} = 2.8 m/s$ $V_i = 2.8" m/s"$ and $s$ $s$ is the distance along the plane:

$- V_{i}^{2} = 2 a s$ $-V_i^2=2as$

$s = \frac{- V_{i}^{2}}{2 a}$ $s =(-V_i^2)/(2a)$

$s = \frac{- {(2.8 m/s)}^{2}}{2 (- 6.43 {m/s}^{2})}$ $s= (-(2.8" m/s")^2)/(2(-6.43" m/s"^2)$

$s \approx 0.61 m$ $s~~ 0.61" m"$

Answer 2 · 2017-07-27T23:54:43

Alternative solution.

Explanation:

Let the block of mass $m$ $m$ move a distance $s$ $s$ up the incline.
Initial energy of block is its kinetic energy $= \frac{1}{2} m v^{2}$ $=1/2mv^2$ ....(1)

As it moves up the incline this energy gets converted in to its potential energy and part energy is spent to overcome force of friction.

$∣ ∣ ∣ {\to F}_{normal} ∣ ∣ ∣ = m g {cos 30}^{\circ}$ $|vecF_"normal"| = mgcos30^@$
where acceleration due to gravity $g = 9.8 m s^{- 2}$ $g=9.8" ms^-2$

We know that ${\to F}_{friction} = μ {\to F}_{normal}$ $vecF_"friction" = muvecF_"normal"$

$:.F_"friction" = mumgcos30^@$

Distance traveled against friction $=s$
Work done against friction $="Force"xx"distance"$
$=mumgcos30^@xxs$ .....(2)

Potential energy of block after traveling distance $s$ along the incline $=mgh=mg(ssin30^@)$ ......(3)

Using Law of conservation of energy we get from (1), (2) and (3)

$1/2mv^2=mumgcos30^@s+mg(ssin30^@)$
$=>(mucos30^@+sin30^@)gs=v^2/2$
$=>s=v^2/(2g(mucos30^@+sin30^@))$

Inserting various values we get
$=>s=(2.8)^2/(2xx9.8(0.18xxsqrt3/2+1/2))$
$=>s=0.61m$ , rounded to two decimal places.

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