A small block of mass \( m \) is attached to a spring with stiffness \( k \) and relaxed length \( L \). The other end of the spring is fastened to a fixed point on a low-friction table. The block slides on the table in a circular path of radius \( R>L \). How long does it take for the block to go around once? \( T= \) Preview My Answers Submit Answers You have attempted this problem 0 times.

To determine the time \( T \) it takes for the block to go around once in a circular path of radius \( R \), we can start by analyzing the forces acting on the block. 1. **Centripetal Force Requirement**: The block is moving in a circular path, so it requires a centripetal force to maintain that motion. The centripetal force \( F_c \) needed for an object of mass \( m \) moving with speed \( v \) in a circle of radius \( R \) is given by: \[ F_c = \frac{mv^2}{R} \] 2. **Spring Force**: The spring exerts a force on the block when it is stretched. The force exerted by the spring when it is stretched to a length \( R \) (where \( R > L \)) is given by Hooke's law: \[ F_s = k(R - L) \] Here, \( R - L \) is the extension of the spring from its relaxed length \( L \). 3. **Equating Forces**: For the block to maintain circular motion, the spring force must provide the necessary centripetal force: \[ k(R - L) = \frac{mv^2}{R} \] 4. **Solving for Speed \( v \)**: Rearranging the equation gives us: \[ mv^2 = k(R - L)R \] \[ v^2 = \frac{k(R - L)R}{m} \] \[ v = \sqrt{\frac{k(R - L)R}{m}} \] 5. **Finding the Period \( T \)**: The period \( T \) of motion is the time it takes to complete one full revolution. The relationship between the speed \( v \) and the period \( T \) is given by: \[ T = \frac{2\pi R}{v} \] Substituting the expression for \( v \): \[ T = \frac{2\pi R}{\sqrt{\frac{k(R - L)R}{m}}} \] Simplifying this expression: \[ T = 2\pi R \sqrt{\frac{m}{k(R - L)R}} \] \[ T = 2\pi \sqrt{\frac{mR}{k(R - L)}} \] Thus, the time it takes for the block to go around once is: \[ \boxed{T = 2\pi \sqrt{\frac{mR}{k(R - L)}}} \]