A 2.97 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 12.9 N. The coefficient of

kinetic friction is 0.126. The acceleration of gravity is 9.8 m/s^2.
Find the speed of the block after it has
moved 2.92 m.

What equation should we use or what? Thanks

To solve this problem, we can use the equations of motion. First, let's analyze the forces acting on the block.

1. The force pulling the block to the right is the applied force, which has a magnitude of 12.9 N.
2. The force of kinetic friction opposes the motion and is given by the equation: \(f_{\text{friction}} = \mu \cdot f_{\text{normal}}\), where \(f_{\text{normal}}\) is the normal force and \(\mu\) is the coefficient of kinetic friction.
3. The normal force is equal to the weight of the block, which is \(f_{\text{normal}} = m \cdot g\), where \(m\) is the mass of the block and \(g\) is the acceleration due to gravity.

Now, let's calculate the forces:

1. Find the weight of the block: \(f_{\text{normal}} = m \cdot g = 2.97\,\text{kg} \cdot 9.8\,\text{m/s}^2 = 29.106\,\text{N}\)
2. Calculate the force of friction: \(f_{\text{friction}} = \mu \cdot f_{\text{normal}} = 0.126 \cdot 29.106\,\text{N} = 3.669756\,\text{N}\)

Next, let's calculate the net force acting on the block:

1. The net force is the difference between the applied force and the force of friction: \(f_{\text{net}} = f_{\text{applied}} - f_{\text{friction}} = 12.9\,\text{N} - 3.669756\,\text{N} = 9.230244\,\text{N}\)

Finally, we can use the equation of motion to find the speed of the block after it has moved 2.92 m:

\[v_f^2 = v_i^2 + 2a \cdot d\]

Since the block starts at rest, the initial velocity (\(v_i\)) is 0. Therefore, the equation simplifies to:

\[v_f = \sqrt{2a \cdot d}\]

where \(a\) is the acceleration and \(d\) is the distance.

Now, let's calculate the acceleration:

1. Use Newton's second law: \(f_{\text{net}} = m \cdot a\) to find \(a\).
\(a = \frac{f_{\text{net}}}{m} = \frac{9.230244\,\text{N}}{2.97\,\text{kg}} = 3.1090864\,\text{m/s}^2\)

Finally, we can calculate the speed of the block after it has moved 2.92 m:

\[v_f = \sqrt{2 \cdot 3.1090864\,\text{m/s}^2 \cdot 2.92\,\text{m}}\]

Calculating this expression will give us the speed of the block after it has moved 2.92 m.

.148 m/s