If it takes total work W to give an object a speed vand kinetic energy K,starting from rest, what will be the object's speed (in terms of v)if we do twice as much work on it, again starting from rest? Round off your result to 1 decimal place. Yanit: square

## Step 1<br />The problem involves the concept of kinetic energy and work done. The kinetic energy (K) of an object is given by the formula:<br />### \(K = \frac{1}{2}mv^2\)<br />where \(m\) is the mass of the object and \(v\) is its velocity.<br /><br />## Step 2<br />The work done (W) on an object is equal to the change in its kinetic energy. This is based on the work-energy principle.<br /><br />## Step 3<br />If we do twice as much work on the object, the new work done (W') is:<br />### \(W' = 2W\)<br /><br />## Step 4<br />The new kinetic energy (K') of the object is:<br />### \(K' = \frac{1}{2}m(v')^2\)<br />where \(v'\) is the new velocity of the object.<br /><br />## Step 5<br />According to the work-energy principle, the work done is equal to the change in kinetic energy. Therefore, we can write:<br />### \(W' = K' - K\)<br /><br />## Step 6<br />Substituting the values of \(W'\), \(K'\), and \(K\) into the equation, we get:<br />### \(2W = \frac{1}{2}m(v')^2 - \frac{1}{2}mv^2\)<br /><br />## Step 7<br />Solving this equation for \(v'\), we get:<br />### \(v' = \sqrt{2}v\)