A7.06 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.

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

be subject to a force andchange its motion

d eras and periods