If it takes total work Wto give an object a speed v and kinetic energy K,starting from rest, what will be the object's kinetic energy (in terms of K) 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 equation \( K = \frac{1}{2}mv^2 \), 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 expressed as \( W = \Delta K = K_{final} - K_{initial} \).<br /><br />## Step 3<br />In this problem, we are given that the initial kinetic energy is zero (since the object starts from rest), and the final kinetic energy is \( K \). Therefore, the work done is equal to \( K \), i.e., \( W = K \).<br /><br />## Step 4<br />If we do twice as much work on the object, the new work done is \( 2W \). Since the initial kinetic energy is still zero, the new final kinetic energy is \( 2K \).<br /><br />## Step 5<br />However, the question asks for the kinetic energy in terms of \( K \), not the absolute value. Therefore, the kinetic energy is \( 2K \), which is twice the original kinetic energy.