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
