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vfiekz [6]
2 years ago
5

One end of an elastic cord is fastened to a steel beam. A metal weight with a mass of 68 kg is attached to the other end of the

cord and dropped from the level of the beam. The weight is traveling at 14 m/s downward when the rest length of the cord is exceeded and the cord begins to stretch and exert an upward force on the metal weight. After 0. 85 seconds, the weight is moving upward with a speed of 11 m/s. What is the average force on the weight during this time period?.
Physics
1 answer:
Vika [28.1K]2 years ago
4 0

Hi there!

We can use IMPULSE to solve.

Recall impulse:

I = \Delta p = m\Delta v = m(v_f - v_i)

Begin by calculating the impulse. Assuming up to be the + direction, and down to be the - direction.

I = 68(11 - (-14)) = 68 \cdot 25 = 1700 kg\frac{m}{s}

Now, we can calculate force using this value:

I = F \cdot t\\\\F = \frac{I}{t}\\\\F = \frac{1700}{0.85} = \boxed{2000 N}

<u>The weight experiences a net force of 2000N UPWARDS.</u>

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What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
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The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

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here We note that the,

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During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
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The maximum height attained is 460 m.

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