A 3.10 kg box that is several hundred meters above the surface of the earth is s

Question

A 3.10 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)=(38.0N/s)t . The box is at rest at t =0. The only forces on the box are the tension in the rope and gravity.

Part A

What is the velocity of the box at 1.00 s ?

vy = -3.67 ms

Part B

What is the velocity of the box at 3.00 s ?

25.8 ms

Part C

What is the maximum distance that the box descends below its initial position?

Part D

At what value of t does the box return to its initial position?

25.8 ms

Explanation / Answer

Here ,

T = 38 * t

net force on the box ,

F = 38 * t - 3.1 * 9.8

F = 38*t - 30.4 N

as a = F/m

a = (38 * t - 3.1 *9.8)/3.1

a = 12.3 t - 9.8

as v = integration( a * dt)

v = integration(12.3 t - 9.8 )

v = 6.15 * t^2 - 9.8 * t

Now , at t = 1 s

v = 6.15 * 1 - 9.8

v = - 3.65 m/s

b)

at t = 3 s

v = 6.15 * 3^2 - 9.8 * 3

v = 26 m/s

C)

for position ,

x = integration(v dt)

x = integration (6.15 * t^2 - 9.8 * t)

x = 2.05 t^3 - 4.9 * t^2

Now , for maximum downwards position ,

v = 0

6.15 * t^2 - 9.8 * t = 0

t = 1.593

position at t = 1.593

x = 2.05 1.593^3 - 4.9 * 1.593^2

x = -4.14 m

the maximum distance that the box descends is 4.14 m

D)

for x = 0

x = 2.05 t^3 - 4.9 * t^2

0= 2.05 t^3 - 4.9 * t^2

t = 2.39 s

block returns to the initial position at t = 2.39 s

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