Answer :
Answer:
The distance moved by the block is 1.34 m
Explanation:
Given;
initial velocity of the block, u = 3 m/s
angle of inclination, θ = 20⁰
The net horizontal force acting on the block;
∑Fx = - mgsinθ
Apply Newton's second law of motion, to determine the constant acceleration of the block.
∑Fx = ma
- mgsinθ = ma
-gsinθ = a
Apply the following Kinematic equation, to determine how far up, the block moved before coming to rest.
[tex]V_f^2 = V_i^2 + 2a(x_f -x_i)\\\\0 = V_i^2 + 2[-gsin \theta(x_f-0)]\\\\0 = V_i^2 - 2gsin \theta(x_f)\\\\ 2gsin \theta(x_f) = V_i^2\\\\x_f = \frac{V_i^2}{2gsin \theta} = \frac{(3)^2}{2 \ \times \ 9.8 \ \times \ sin(20)} = 1.34 \ m[/tex]
Therefore, the distance moved by the block is 1.34 m