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SpyIntel [72]
3 years ago
7

"A 3 kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30 degrees. The crate starts from rest a

t the top. Determine the speed of the crate at the bottom of the ramp if the coefficient of kinetic friction between the crate and ramp is 0.19"
Physics
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

v = 2.57 m /sec

Explanation: See Annex Free Body Diagram

From free body diagram and Newton´s second law we have

There is not movements in the y axis direction

cos 30°  =  √3/2       sin 30°  =  1/2

We have  P  = mg   =  3 Kg  *  9.8 m/sec²

P  =   29.4  Kg*m/ sec²      P  =  29.4 [N]

Py  =  P * cos 30°    Py =  29.4 [N] * √3/2   ⇒    Py = 25.43 [N]

Px  =   P * sin 30°    Px =  29.4 [N] * 1/2      ⇒     Px = 14.7  [N]

∑ F   =  m* a         ⇒    ∑ Fy   =  0       ∑ Fx  =  m *a

∑ Fy   =  Fn  -  Py   =  0         Py   = P*cos30°       Py = 25.43 [N]

Fn  =  25.43 [N]

Fr  =  μk * Fn      ⇒   Fr  =  0.19 * 25.43   ⇒ Fr  =  4.83 [N]    

Now

∑ Fx  =  m *a       mg sin30° - Fr =  m*a    ⇒   Px  - Fr  = m*a

14.7 [N]   -  4.83 [N]   =  3 [kg] * a       ⇒   9.87 /3   = a [m /sec²]

a = 3.29 [m/sec²]

From uniformly accelerated movement

distance  =  x₀  + V₀*t ± at²/2     but  x₀   and  V₀    =  0

Then

d = ( 1/2 )*a*t²     ⇒  1 [m]  * 2  =  3.29 [m/sec²] * t²

t  =  0.78 sec

And finally

v =  a*t      ⇒   v  =  3.29 *(.78)     ⇒   v = 2.57 m /sec

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A block (6 kg) initially compresses spring #1 (k = 2000 N/m) by 60 cm from its equilibrium point. When the block is released, it
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Answer:

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Explanation:

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        W_{fr}= Ef - E₀

Let's look for the energies

Initial

        E₀ = Ke = ½ k₁  x₁²

Final, this is just before starting to compress the spring

        Ef = Ke = ½ m v²

The work of the rubbing force is

       W_{fr}= -fr x

Let's write Newton's second law the y axis

       N-W = 0

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      fr = μ N

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Let's replace

      -μ mg x = ½ m v² - ½ k₁ x₁²

       v² = 2/m (½ k₁ x1₁² -μ mg x)

       v² = 2/6  (½ 2000 0.6²2 - 0.5  6  9.8 1) = 1/3 (360 - 29.4)

       v = 3.13 m / s

With this value we calculate the energy of the block

       K = ½ m v²

       K = ½  6  3.13²

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Calculate eenrgy of the spring ke 1

      Ke = ½ k₁ x₁²

      Ke = ½ 2000 0.60²

      Ke = 360 J

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