Aglider with mass m = 0.200 kg sits on a frictionless, horizontal air track. it is connected to a spring of negligible mass and force constant k = 5.00 n/m. you pull on the glider to stretch the spring 0.100 m, and then release the glider. the glider begins to move back toward the equilibrium position (x = 0). what is its speed when x = 0.0800 m?

0.400 m/s

0.300 m/s

0.016 m/s

0.025 m/s

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

answer: electrons