Ask Question
13 February, 12:58

A 7.06 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant, horizontal force of 12.1 N. Find the speed of the block after it has moved 3 m. Answer in units of m/s.

+3
Answers (1)
  1. 13 February, 13:42
    0
    The speed of the block = 3.20 m/s

    Explanation:

    Speed: This is defined as the rate of change of distance. The S. I unit of speed is m/s.

    Force = Mass * Acceleration

    F = ma

    making a the subject of the equation,

    a = F/m ... Equation 1

    Given: F = 12.1 N, m = 7.06 kg.

    Substituting these values into equation 1

    a = 12.1/7.06

    a = 1.71 m/s²

    Using Newton's equation of motion,

    v² = u² + 2as ... Equation 2

    Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

    Given: u = 0 (the block was initially at rest), a = 1.71 m/s², s = 3 m

    Substituting these values into equation 2,

    v² = 0² + 2 (1.71) (3)

    v² = 10.26

    √v² = √10.26

    v = 3.20 m/s

    Therefor the speed of the block = 3.20 m/s
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A 7.06 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant, horizontal force of 12.1 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers