A 5.9 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 2.6 m.

To find the speed of the block after it has moved a certain distance, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

The work done on the block by the applied force can be calculated using the formula:

Work = Force * Distance * cos(theta)

In this case, the force is 12.1 N and the distance is 2.6 m. Since the force and the displacement are both horizontal, the angle theta between them is 0 degrees, so the cosine of theta is 1.

Plugging in the values, we get:

Work = 12.1 N * 2.6 m * cos(0) = 31.46 J

The work done on the block is equal to the change in its kinetic energy, so we can use this information to find the final speed of the block.

The change in kinetic energy is given by the formula:

Change in kinetic energy = Kinetic energy final - Kinetic energy initial

Since the block starts from rest, its initial kinetic energy is zero. Therefore, we have:

Change in kinetic energy = Kinetic energy final - 0

The change in kinetic energy is equal to the work done on the block. So, we have:

31.46 J = Kinetic energy final - 0

Therefore, the final kinetic energy of the block is equal to 31.46 J. The formula for kinetic energy is:

Kinetic energy = 0.5 * mass * (velocity)^2

Rearranging the formula, we can solve for velocity:

Velocity = square root(2 * Kinetic energy / mass)

Plugging in the values, we have:

Velocity = square root(2 * 31.46 J / 5.9 kg) ≈ 2.69 m/s

Therefore, the speed of the block after it has moved 2.6 m is approximately 2.69 m/s.

F = m a

12.1 = 5.9 a
solve for a

d = (1/2) a t^2
2.6 = (1/2) a t^2
we know a, solve for t

v = a t