CALC? 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) = (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. (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?
Solution 63P Step 1: Step 2: a) he resultant force along upward direction, F = T - mg Where, T - Tension on the string mg - weight of the box which is acting downwards. That is, ma = T - mg Divide the equation by using “m”, which is the mass of the box a = (T/m) - g But, we know that, T = (36 N/s) t Therefore, the equation becomes, a = (36 /m) (N/kg s) t - g We know that, the acceleration, a = dv/dt Therefore, dv = a dt That is, v = a dt = [(36 /m) (N/kg s) t - g]dt v = (3 /m) (N/kg s) t dt - g dt v =(36 /m) (N/kg s) t dt - g d 2 v = (36 /m) (N/kg s) (t /2) - gt + C When t = 0, we know that, v = 0 2 Therefore, v = (36 /m) (N/kg s) (t /2) - gt i) So, we have to find the velocity when t = 1 s 2 2 That is, v =1(36 /m) (N/kg s) (1 /2) s ]- (g. 1 s) v1 [(36 /2m) - 9.8] m/s 2 Since, g = 9.8 m/s and m = 3 kg, v1 (36/ 6) - 9.8 = 6 - 9.8 = - 3.8 m/s ii) When, t = 2 s 2 2 v2 [(36 /m) (N/kg s) (2 /2) s ]- (g. 2 s) v2 [(36×2 /m) - (9.8×2) ] m/s v2 [(72 /3) - (19.6) ] m/s v2 24 - 19.6 = 4.4 m/s