A 0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of
1 answer:
Answer:
Explanation:
k = spring constant of the spring = 40 Nm⁻¹
A = amplitude of the simple harmonic motion = 4 cm = 0.04 m
m = mass of the block attached to spring = 0.20 kg
w = angular frequency of the simple harmonic motion
Angular frequency of the simple harmonic motion is given as
= Speed of the block as it pass the equilibrium point
Speed of the block as it pass the equilibrium point is given as
You might be interested in
Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.
