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) \( 5.35 d \)

The kinetic energy given to the mass is equal to the elastic potential energy stored in the spring. For a spring compressed by a distance d, the potential energy is   PE = (1/2) k d². When the spring is released, this energy becomes kinetic energy (KE) of the mass m:   (1/2) k d² = (1/2) m v²      (1) Now, suppose we want the mass to have twice the original speed, i.e., 2v. The kinetic energy for the new launch would be   KE_new = (1/2) m (2v)² = (1/2) m (4v²) = 2 m v². We want this KE_new to come from the spring potential energy when compressed by an unknown distance d_new:   (1/2) k d_new² = 2 m v²      (2) From equation (1), we have:   (1/2) m v² = (1/2) k d². Substituting m v² from equation (1) into equation (2):   (1/2) k d_new² = 2 × [(1/2) k d²]  → (1/2) k d_new² = k d². Canceling (1/2) k from both sides:   d_new² = 2 × 2 d²    [Actually, correctly: dividing both sides by (1/2)k gives:]   d_new² = 4 d² Taking the square root of both sides:   d_new = 2 d. Thus, to double the speed, you must compress the spring a distance of 2.00 d. The correct answer is: 2) 2.00 d. (Note: Among the provided choices, 2.00 d corresponds to option 3.)