A 3.2 kg block initially at rest is pulled to the

right along a horizontal, frictionless surface
by a constant, horizontal force of 19.4 N.
Find the speed of the block after it has
moved 2.5 m.
Answer in units of m/

Calculate the acceleration a, using

a = F/m.
Then use V = sqrt(2aX)

6.06

To find the speed of the block after it has moved 2.5m, we can use Newton's second law of motion. The formula for Newton's second law is:

F = m * a

where F is the net force applied to the object, m is the mass of the object, and a is the acceleration of the object.

In this case, the net force applied to the object is 19.4N, and the mass of the block is 3.2kg. We need to find the acceleration of the block in order to determine its final speed.

We can find the acceleration using the following equation:

a = F / m

Substituting the given values, we have:

a = 19.4N / 3.2kg

a ≈ 6.06 m/s^2

Now, we can use the kinematic equation to find the final speed of the block. The kinematic equation is:

v^2 = u^2 + 2a * s

where v is the final velocity, u is the initial velocity (which is 0 since the block is initially at rest), a is the acceleration, and s is the displacement.

Substituting the given values, we have:

v^2 = 0^2 + 2 * 6.06 m/s^2 * 2.5m

v^2 ≈ 30.3 m^2/s^2

Taking the square root of both sides gives us:

v ≈ √30.3 m^2/s^2

v ≈ 5.5 m/s

Therefore, the speed of the block after it has moved 2.5m is approximately 5.5 m/s.