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zalisa [80]
3 years ago
9

A 4.76 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

d of the rope, and the height of the crate above its initial position is given by y(t)=(2.80m/s)t +(0.61 m/s3 )t3
What is the magnitude of the force F when 3.71 s ?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0

Answer:

The magnitude of the force is 64.634 newtons.

Explanation:

According to the statement, the crate is a constant mass system, whose upward force is described by the following expression:

F(t) = m\cdot \ddot{y} (t) (1)

Where:

F(t) - Force, in newtons.

m - Mass, in kilograms.

\ddot {y}(t) - Acceleration, in meters per square second.

The function acceleration is obtained by deriving the function position twice in time:

\dot y (t) = 2.80 + 1.83\cdot t^{2} (2)

\ddot y(t) = 3.66\cdot t (3)

And we expand (1) by applying (3):

F(t) = 3.66\cdot m \cdot t

Where t is the time, in seconds.

If we know that m = 4.76\,kg and t = 3.71\,s, then the magnitude of the force is:

F = 3.66\cdot (4.76)\cdot (3.71)

F = 64.634\,N

The magnitude of the force is 64.634 newtons.

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