(b) A horizontal spring with force constant 500N/m is attached at one end with a mass of 5 kg which is kept on a horizontal frictionless table. The other end of the string is attached to a fixed support. If the mass is pulled 20 cm from its equilibrium position along the table and released, calculate the velocity the acceleration and the potential and kinetic energies of the mass when it has moved 5 cm towards the equilibrium position from its initial position (o marks)

To solve this problem, we need to use the principles of simple harmonic motion (SHM) and energy conservation.<br /><br />1. **Spring Force Constant and Mass:**<br /> - Spring constant, \( k = 500 \, \text{N/m} \)<br /> - Mass, \( m = 5 \, \text{kg} \)<br /><br />2. **Initial Displacement:**<br /> - Initial displacement from equilibrium, \( x_0 = 20 \, \text{cm} = 0.2 \, \text{m} \)<br /><br />3. **Position at 5 cm towards Equilibrium:**<br /> - New displacement from equilibrium, \( x = 15 \, \text{cm} = 0.15 \, \text{m} \)<br /><br />4. **Velocity Calculation:**<br /> - The velocity in SHM can be calculated using energy conservation:<br /> \[<br /> \frac{1}{2} k x_0^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2<br /> \]<br /> Solving for \( v \):<br /> \[<br /> v = \sqrt{\frac{k}{m} (x_0^2 - x^2)}<br /> \]<br /> Substituting the values:<br /> \[<br /> v = \sqrt{\frac{500}{5} (0.2^2 - 0.15^2)} = \sqrt{100 (0.04 - 0.0225)} = \sqrt{100 \times 0.0175} = \sqrt{1.75} \approx 1.32 \, \text{m/s}<br /> \]<br /><br />5. **Acceleration Calculation:**<br /> - Acceleration in SHM is given by:<br /> \[<br /> a = -\frac{k}{m} x<br /> \]<br /> Substituting the values:<br /> \[<br /> a = -\frac{500}{5} \times 0.15 = -100 \times 0.15 = -15 \, \text{m/s}^2<br /> \]<br /><br />6. **Potential Energy Calculation:**<br /> - Potential energy stored in the spring:<br /> \[<br /> PE = \frac{1}{2} k x^2 = \frac{1}{2} \times 500 \times (0.15)^2 = 0.5 \times 500 \times 0.0225 = 5.625 \, \text{J}<br /> \]<br /><br />7. **Kinetic Energy Calculation:**<br /> - Kinetic energy can be found using energy conservation:<br /> \[<br /> KE = \frac{1}{2} k x_0^2 - PE = \frac{1}{2} \times 500 \times (0.2)^2 - 5.625 = 10 - 5.625 = 4.375 \, \text{J}<br /> \]<br /><br />In summary:<br />- Velocity when the mass has moved 5 cm towards equilibrium: \( v \approx 1.32 \, \text{m/s} \)<br />- Acceleration at that point: \( a = -15 \, \text{m/s}^2 \)<br />- Potential energy: \( PE = 5.625 \, \text{J} \)<br />- Kinetic energy: \( KE = 4.375 \, \text{J} \)