Answer:
a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s
Explanation:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
We write the equation of the forces acting on the mass.
So, T - mg = ma where T = tension in vertical rope = (36.0 N/s)t, m = mass of box = 3.50 kg, g = acceleration due to gravity = 9.8 m/s² and a = acceleration of box = dv/dt where v = velocity of box and t = time.
So, T - mg = ma
T/m - g = a
dv/dt = T/m - g
dv/dt = (36.0 N/s)t/3.50 kg - 9.8 m/s²
dv/dt = (10.3 m/s²)t - 9.8 m/s²
dv = [(10.3 m/s²)t - 9.8 m/s²]dt
Integrating, we have
∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt
∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt
v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C
v = (5.15 m/s³)t² - (9.8 m/s²)t + C
when t = 0, v = 0 (since at t = 0, box is at rest)
So,
0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C
0 = 0 + 0 + C
C = 0
So, v = (5.15 m/s³)t² - (9.8 m/s²)t
i. What is the velocity of the box at t = 1.00 s,
v = (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)
v = 5.15 m/s - 9.8 m/s
v = -4.65 m/s
ii. What is the velocity of the box at t = 3.00 s,
v = (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)
v = 15.45 m/s - 29.4 m/s
v = -13.95 m/s
b. What is the maximum distance that the box descends below its initial position?
Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t
dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt
Integrating, we have
∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt
∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt
∫dy = ∫(5.15 m/s³)t³/3dt - ∫(9.8 m/s²)t²/2dt
y = (1.72 m/s³)t³ - (4.9 m/s²)t² + C'
when t = 0, y = 0.
So,
0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'
0 = 0 + 0 + C'
C' = 0
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
The maximum distance is obtained at the time when v = dy/dt = 0.
So,
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0
(5.15 m/s³)t² - (9.8 m/s²)t = 0
t[(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or (5.15 m/s³)t = (9.8 m/s²)
t = 0 or t = (9.8 m/s²)/(5.15 m/s³)
t = 0 or t = 1.9 s
Substituting t = 1.9 s into y, we have
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²
y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)
y = 11.798 m - 17.689 m
y = -5.891 m
y ≅ - 5.89 m
So, the maximum distance that the box descends below its initial position is 5.89 m
c. At what value of t does the box return to its initial position?
The box returns to its original position when y = 0. So
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
0 = (1.72 m/s³)t³ - (4.9 m/s²)t²
(1.72 m/s³)t³ - (4.9 m/s²)t² = 0
t²[(1.72 m/s³)t - (4.9 m/s²)] = 0
t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0
t = √0 or (1.72 m/s³)t = (4.9 m/s²)
t = 0 or t = (4.9 m/s²)/(1.72 m/s³)
t = 0 or t = 2.85 s
So, the box returns to its original position when t = 2.85 s
is the stretch of the spring with respect to its equilibrium position. Using the data, we find
