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Theresab2021
12.12.2019 •
Physics
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
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Ответ:
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
Ответ:
1) 1,157*10⁻⁵ day; 2) 604800 sec.; 3) 0.5 m.
Explanation:
1) 1 second into one day:
one day consists of 24 hours, it means 24*3600=86400 seconds. Then 1 second is 1/86400 day or 1,157*10⁻⁵ day.
2) 1 week into seconds:
1 day consists of 3600*24=86400 seconds, then the week is 7*86400=604800 seconds.
3) 50 cm into meter:
100 cm is one meter, then 50 cm is 0.5 meter.
PS. note, there is not the only way to perform conversion.