Question

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 end 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 ?

Answer 1

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.