Question

A block is given an initial velocity of 3.00 m/s up a frictionless incline of angle 20 degrees. How far up the incline does the block slide

Answer 1

The distance moved by the block is 1.34 m

Step-by-step 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.

`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 = (V_i^2)/(2gsin \theta) = ((3)^2)/(2 \ * \ 9.8 \ * \ sin(20)) = 1.34 \ m`

Therefore, the distance moved by the block is 1.34 m