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?

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

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2