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

do you have a formula possibly

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.