Question

"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 at 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"

Answer 1

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