A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance \( d \) and is used to launch a mass \( m \) along the frictionless surface. What compression of the spring would result in the mass attaining double the speed received in the above situation? 1) \( 1.41 d \) 2) \( 1.73 d \) 3) \( 2.00 d \) 4) \( 4.00 d \) 5) \( 5.35 d \)

To make the mass \( m \) reach double the speed, we can use the concept of energy. The potential energy stored in the spring when compressed a distance \( d \) can be expressed with Hooke's Law as \( \frac{1}{2} k d^2 \), where \( k \) is the spring constant. The kinetic energy when the mass exits the spring is given by \( \frac{1}{2} mv^2 \). If we want the mass to attain double the original speed, \( 2v \), its kinetic energy becomes \( \frac{1}{2} m (2v)^2 = 2mv^2 \). The ratio of energies gives us \( 2mv^2/\frac{1}{2} k d^2 = 4 \) leading to the new compression being \( d' = 2d \) or even more specifically, we need to double the original energy, which corresponds to \( 4d^2 \times k/2 = 4\%, \) resulting in a compression of \( 2d \). So, looking at the problem, we can deduce that the answer is \( 4.00 d \). In summary, the answer is option 4: \( 4.00 d \).