A 4.9 kg block initially at rest is pulled to the

right along a horizontal, frictionless surface
by a constant, horizontal force of 14 N.
Find the speed of the block after it has

Incomplete.

F=MA>

sqrt(2AX)
distance is X

23

To find the speed of the block after it has been pulled, we can use the concept of work-energy theorem.

First, we need to find the work done on the block by the applied force. The work done (W) is equal to the product of the force (F) and the displacement (s) in the direction of the force. In this case, the force is 14 N and the displacement is unknown. However, we know that the block starts from rest and we want to find its speed after being pulled. This means the displacement is related to the final velocity. If we assume the initial velocity is 0, then we have:

W = F * s = (14 N) * s

According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. The initial kinetic energy of the block is 0 since it is at rest, and the final kinetic energy is (1/2) * m * v^2, where m is the mass of the block and v is its final velocity.

So, we can set up the equation as follows:

W = (1/2) * m * v^2 - (1/2) * m * 0^2

Since the block is initially at rest, the initial kinetic energy term is 0.

Substituting the value of W:

(14 N) * s = (1/2) * m * v^2

Now, we need to find the displacement (s). Since the block is pulled in a horizontal direction, the displacement is the same as the distance covered by the block. We don't have enough information in the problem statement to determine the distance. Therefore, we cannot find the exact speed of the block after it has been pulled using this method.

However, if we assume that the block is pulled over a distance of d, then we can substitute the known values into the equation and solve for v:

(14 N) * d = (1/2) * (4.9 kg) * v^2

Simplifying the equation:

14 N * d = 2.45 kg * v^2

Now, we can solve the equation for v:

v^2 = (14 N * d) / (2.45 kg)

v = sqrt((14 N * d) / (2.45 kg))

Please provide the distance over which the block is pulled, and I can calculate the speed for you.