Question to (1 point) Judy has a mass of 35Kg and is roller skating with a volocity of 4m/s. Jimmy skates up behind her and pushes her, increasing her volocity to 5.5m/s How much work did Jimmy accomplish? a 5323. b 249.4. 184J d 1.5 J

## Step 1<br />The work done on an object is equal to the change in its kinetic energy. This is based on the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.<br /><br />## Step 2<br />The formula for kinetic energy is given by:<br />### \( KE = \frac{1}{2}mv^2 \)<br />where \( m \) is the mass of the object and \( v \) is its velocity.<br /><br />## Step 3<br />Calculate the initial kinetic energy of Judy. Substituting the given values into the formula, we get:<br />### \( KE_{initial} = \frac{1}{2} \times 35 \, kg \times (4 \, m/s)^2 = 280 \, J \)<br /><br />## Step 4<br />Calculate the final kinetic energy of Judy after Jimmy pushed her. Substituting the given values into the formula, we get:<br />### \( KE_{final} = \frac{1}{2} \times 35 \, kg \times (5.5 \, m/s)^2 = 538.75 \, J \)<br /><br />## Step 5<br />The work done by Jimmy is the difference between the final and initial kinetic energies. So, we subtract the initial kinetic energy from the final kinetic energy to get the work done by Jimmy.<br />### \( Work = KE_{final} - KE_{initial} = 538.75 \, J - 280 \, J = 258.75 \, J \)