A 3.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to v (t) = (2.00m/s2) t + (0.600m/s3) t2.

Question

Rory Lynch

Physics

5 December, 17:16

A 3.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to v (t) = (2.00m/s2) t + (0.600m/s3) t2.

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Aracely Oneal · Answer 1

T = 35.4 + 3.60t

Substitute the time in seconds to get the value for t.

Explanation:

Since the box has a velocity function V (t) which changes with time the box is accelerating

a = dv/dt = 2.00 + 1.20t (obtained by differentiating with respect to t.)

There are two forces acting on the box which are the tension in the string and the weight of the string mg.

From Newtown's second law,

T - mg = ma

T = ma + mg

Substituting the value for a into the equation,

T = m (2.00 + 1.20t + 9.8)

= m (11.8 + 1.20t)

Given m = 3.00kg

T = 3 (11.8 + 1.20t)

= 35.4 + 3.60t