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Kruka [31]
3 years ago
13

A 2.00-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. Part A What is the tension in the rope when the velocity of the box is 15.0 m/s ?
Physics
1 answer:
Alex3 years ago
4 0

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

      T - W = m a

The weight

      W = mg

The acceleration can be found by derivatives

     a = dv / dt

     v = 2 t + 0.6 t²

     a = 2 + 0.6 t

We replace

      T - mg = m (2 + 0.6t)

      T = m (g + 2 + 0.6 t)               (1)

Let's look for the time for the speed of 15 m / s

       15 = 2 t + 0.6 t²

       0.6 t² + 2 t - 15 = 0

We solve the second degree equation

        t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6

        t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

       t = 3.6 s

Let's calculate from equation 1

       T = 2.00 (9.8 + 2 + 0. 6  3.6)

        T = 27.92 N

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3 years ago
An object is dropped from 600 m. If it is initially at rest and achieves a top speed of 45 m/s just as it hits the ground, what
Brums [2.3K]

Answer:

1.68m/s^2

Explanation:

using V^2=U^2+2×a×s formula.

If,

(Final Velocity)V=45

(Initial Velocity)U=0 because it starts from rest

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(acceleration)a= ?

now,

using formula,

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Hope it works !!!

3 0
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Read 2 more answers
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