A 1.5-kg physics textbook is initially at rest on a steel table.The textbook is then pushed with a constant force of 3.0 N. Friction with a magnitude of 1.8 N is exerted on the moving book by the surface of the table.Use the equations for work and kinetic energy to calculate the final velocity of the textbook after it has been pushed 1.0 meter across the table.

## Step 1<br />The problem involves the principles of work and kinetic energy. The work done on an object is equal to the change in its kinetic energy. The work done by a force is calculated by the formula:<br />### \(W = F \cdot d \cdot \cos(\theta)\)<br />where \(W\) is the work done, \(F\) is the force, \(d\) is the distance, and \(\theta\) is the angle between the force and the direction of motion.<br /><br />## Step 2<br />In this case, the force is applied in the same direction as the motion, so \(\theta = 0\) and \(\cos(\theta) = 1\). Therefore, the work done by the applied force is:<br />### \(W_{\text{applied}} = F_{\text{applied}} \cdot d = 3.0 \, \text{N} \cdot 1.0 \, \text{m} = 3.0 \, \text{J}\)<br /><br />## Step 3<br />The work done by the frictional force is negative because it opposes the motion. The work done by the frictional force is:<br />### \(W_{\text{friction}} = F_{\text{friction}} \cdot d = 1.8 \, \text{N} \cdot 1.0 \, \text{m} = 1.8 \, \text{J}\)<br /><br />## Step 4<br />The total work done on the textbook is the sum of the work done by the applied force and the work done by the frictional force:<br />### \(W_{\text{total}} = W_{\text{applied}} - W_{\text{friction}} = 3.0 \, \text{J} - 1.8 \, \text{J} = 1.2 \, \text{J}\)<br /><br />## Step 5<br />The change in kinetic energy of the textbook is equal to the total work done on it:<br />### \(\Delta KE = W_{\text{total}} = 1.2 \, \text{J}\)<br /><br />## Step 6<br />The kinetic energy of an object is given by the formula:<br />### \(KE = \frac{1}{2}mv^2\)<br />where \(m\) is the mass of the object and \(v\) is its velocity.<br /><br />## Step 7<br />We can rearrange the formula to solve for the velocity:<br />### \(v = \sqrt{\frac{2 \cdot \Delta KE}{m}}\)<br /><br />## Step 8<br />Substituting the given values into the formula, we get:<br />### \(v = \sqrt{\frac{2 \cdot 1.2 \, \text{J}}{1.5 \, \text{kg}}} = 1.1 \, \text{m/s}\)