Answer:
0·300 m/s
Explanation:
Given
Mass of the glider = 0·2 kg
Spring constant = k = 5 N/m
At x = 0·1 m, the glider is released, it means that initial velocity of the glider is zero
∴ Total Mechanical energy = potential energy
Equilibrium position is at x = 0 cm
Potential energy in case of spring = 0·5 × k × x²
where k is the spring constant
x is the distance from equilibrium position
Initially potential energy = 0·5 × 5 × (0·1 - 0)² = 0·025 J
As there is no dissipative force, therefore mechanical energy of the glider remains constant
At x = 0·08 m
Mechanical energy of the glider = Kinetic energy + Potential energy
Let the velocity of the glider at x = 0·08 m be v m/s
Kinetic energy at this instant = 0.5 × m × v² = 0·5 × 0·2 × v² J
Potential energy at this instant = 0·5 × k × x² = 0·5 × 5 × 0·08² = 0·016 J
Mechanical energy = 0·5 × 0·2 × v² + 0·016
As mechanical energy is constant
0·025 = 0·5 × 0·2 × v² + 0·016
0·009 = 0·1 × v²
∴ v = 0·300 m/s
∴ Speed when x = 0.0800 m is 0·300 m/s
