A 3.00-kg box that is several hundred meters above the earth's surface 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 results in a tension in the rope of T (t) = (38.0 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. A) What is the velocity of the box at t = 1.00 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? D) What is the velocity of the box at t = 3.00 s?

Question

Keon Avery

Physics

19 December, 10:55

A 3.00-kg box that is several hundred meters above the earth's surface 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 results in a tension in the rope of T (t) = (38.0 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. A) What is the velocity of the box at t = 1.00 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? D) What is the velocity of the box at t = 3.00 s?

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Answers (1)

Know the Answer?

Alonso Sheppard · Answer 1

A) 3.48m/s

B) 3.92m

C) 2.32m

D 23.33m/s

Explanation:

ma (t) = mg-At

a (t) = g - (At/m)

V (t) = integrala (T) dT = gt - (At^2/2m)

Initial x coordinator of the box is zero

X (t) = integralV (t) dt = 1/2gt^2 - (At^3/6m)

a) V = (9.81*1) - (38*1^2/2*3)

V = 9.81-6.33 = 3.48m/s

b) - AT^2/2m + gT = 0

T=2mg/A = (2*3*9.81) / 38

T = 1.549m

X (T) = (1/2*9.81*1.549^3) - (38*1.549^3/6*3)

X (T) = 11.768 - (141.23/18) = 11.768 - 7.85 = 3.92m

C) 1/2gT''^2 - AT''^3/6m = 0

The only non trivial solution is T'' = 3mg/A

T = (3*3*9.81) / 38 = 2.32m

D) V = 9.81*3) - (38*3^2/6)

V = 29 - 5.667 = 23.33m/s