Explanation:
weight =0.454 × 9.8=4.4492N
Answer:
A. Area under force-time graph & Area under force-displacement graph
Explanation:
To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.
Impulse = Force x time
Work done = Force x displacement
When we plot a graph of force and time, the area under it is the impulse.
When a graph of force and displacement is plotted, the area under is the work done.
Answer:
1) k = 52 N/m
2) E = 1.0 J
3) ω = 8.1 rad/s
4) v = 1.4 m/s
Though asked for a velocity, we can only supply magnitude (speed) because we don't have enough information to determine direction.
If it happens to be the first time it is at y = - 10 cm after release, the velocity is upward.
Explanation:
Assuming the initial setup is after all transients are eliminated.
kx = mg
k = mg/x = 0.8(9.8) / 0.15
k = 52.26666.... ≈ 52 N/m
E = ½kA² = ½(52)(0.20²) = 1.045333... ≈ 1.0 J
ω = √(k/m) = √(52 / 0.8) = 8.0829... ≈ 8.1 rad/s
½mv² = ½kA² - ½kx²
v = √(k(A² - x²)/m) = √(52(0.20² - 0.10²)/0.8) = 1.39999... ≈ 1.4 m/s
