(b) A horizontal spring with force constant 1000N/m is attached at one end with a mids of 20 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 10 cm from its equilibrium position along the table and released, calculate the following for the mass when it is 5 cm from the equilibrium position:I (i) the velocity (ii) the acceleration . (iii) the potential crergy (iv) the kinetic energy

To solve this problem, we will use the principles of simple harmonic motion (SHM) and energy conservation. The mass-spring system follows Hooke's Law, and the total mechanical energy is conserved.<br /><br />Given:<br />- Spring constant \( k = 1000 \, \text{N/m} \)<br />- Mass \( m = 20 \, \text{kg} \)<br />- Initial displacement \( x_0 = 10 \, \text{cm} = 0.1 \, \text{m} \)<br />- Displacement at the point of interest \( x = 5 \, \text{cm} = 0.05 \, \text{m} \)<br /><br />**(i) Velocity:**<br /><br />The total mechanical energy in the system is conserved and is given by:<br /><br />\[ E = \frac{1}{2} k x_0^2 \]<br /><br />At any position \( x \), the total energy is the sum of kinetic energy (KE) and potential energy (PE):<br /><br />\[ E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 \]<br /><br />Setting these equal gives:<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 />\[ \frac{1}{2} m v^2 = \frac{1}{2} k x_0^2 - \frac{1}{2} k x^2 \]<br /><br />\[ m v^2 = k (x_0^2 - x^2) \]<br /><br />\[ v^2 = \frac{k}{m} (x_0^2 - x^2) \]<br /><br />\[ v = \sqrt{\frac{k}{m} (x_0^2 - x^2)} \]<br /><br />Substitute the values:<br /><br />\[ v = \sqrt{\frac{1000}{20} (0.1^2 - 0.05^2)} \]<br /><br />\[ v = \sqrt{50 (0.01 - 0.0025)} \]<br /><br />\[ v = \sqrt{50 \times 0.0075} \]<br /><br />\[ v = \sqrt{0.375} \]<br /><br />\[ v \approx 0.612 \, \text{m/s} \]<br /><br />**(ii) Acceleration:**<br /><br />Acceleration \( a \) in SHM is given by:<br /><br />\[ a = -\frac{k}{m} x \]<br /><br />Substitute the values:<br /><br />\[ a = -\frac{1000}{20} \times 0.05 \]<br /><br />\[ a = -50 \times 0.05 \]<br /><br />\[ a = -2.5 \, \text{m/s}^2 \]<br /><br />**(iii) Potential Energy:**<br /><br />Potential energy \( PE \) at displacement \( x \) is:<br /><br />\[ PE = \frac{1}{2} k x^2 \]<br /><br />Substitute the values:<br /><br />\[ PE = \frac{1}{2} \times 1000 \times (0.05)^2 \]<br /><br />\[ PE = 500 \times 0.0025 \]<br /><br />\[ PE = 1.25 \, \text{J} \]<br /><br />**(iv) Kinetic Energy:**<br /><br />Kinetic energy \( KE \) can be found using the total energy or directly from the velocity:<br /><br />\[ KE = \frac{1}{2} m v^2 \]<br /><br />Using the energy conservation approach:<br /><br />\[ KE = E - PE \]<br /><br />\[ KE = \frac{1}{2} k x_0^2 - \frac{1}{2} k x^2 \]<br /><br />\[ KE = \frac{1}{2} \times 1000 \times (0.1^2 - 0.05^2) \]<br /><br />\[ KE = 500 \times (0.01 - 0.0025) \]<br /><br />\[ KE = 500 \times 0.0075 \]<br /><br />\[ KE = 3.75 \, \text{J} \]<br /><br />Thus, the answers are:<br />- Velocity: \( v \approx 0.612 \, \text{m/s} \)<br />- Acceleration: \( a = -2.5 \, \text{m/s}^2 \)<br />- Potential Energy: \( PE = 1.25 \, \text{J} \)<br />- Kinetic Energy: \( KE = 3.75 \, \text{J} \)