A 3.50 kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of T(t) = (36.0 N/s)t. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity. Required:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
b. What is the maximum distance that the box descends below its initial position?
c. At what value of t does the box return to its initial position?

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