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Dmitry_Shevchenko [17]
3 years ago
8

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
Physics
1 answer:
daser333 [38]3 years ago
6 0

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.

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

Therefore, the distance moved by the block is 1.34 m

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