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l Sprint LTE 7:54 AM 1 8396 Touch to return to Navigation session.masteringphysics.com C Problem 5.69 MasteringPhysics 6 of 13 Problem 5.69 A 2.80 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) = (36.0N/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? Express your answer with the appropriate units. v,--3.37 Submit Give Up Correct

Explanation / Answer

Net force F = 36*t - m*g

a = F/m = [36t - mg]/m = 36t/m - g

v = a*dt = 36t²/(2m) - gt = 36t²/(2 x 2.8) – gt = 6.429t² - gt......(The integration constant is zero because v = 0 @ t = 0.)

d = v*dt = 6.429t³/3 - gt² = 2.143t³ - 4.905t² [since g = 9.81 m/s²]

1. v1 = 6.429(1)² - (9.81)(1) = -3.381 m/s (ans)

2. v3 = 6.429(3)² - (9.81)(3) = +28.431 m/s

3.

dmax is obtained when v = 0; From the eq for v (v set to 0), t = g/6.429 = 1.523 sec

From the eq for d, dmax = 2.143(1.523)³ - 4.905(1.523)² = -3.764 m (ans)

4. Set d = 0 in the distance eq: 2.143t³ - 4.905t² = 0

t = 4.905/2.143 = 2.289 sec (ans)

Hope this helps :)

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