Question

A 6.50 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+(0.610m/s3)t3. What is the magnitude of F when t = 4.00 s?

Answer 1

Answer:

$F=158.86 N$

Step-by-step explanation:

We are given that

Mass of crate=m=6.5 Kg

Height,$y(t)=(2.80m/s)t+(0.610m/s^3)t^3$

We have to find the magnitude of F when t=4 s

Differentiate w.r.t t

$v=\frac{dy}{dt}=2.8+3(0.61)t^2$

Again differentiate w.r.t t

$a=\frac{dv}{dt}=3\times 2(0.61)t$

$a=3.66tm/s^3$

Substitue t=4

$a=3.66\times 4=14.64m/s^2$

$F-mg=ma$

Where $g=9.8m/s^2$

Substitute the values

$F-6.5\times 9.8=6.5\times 14.64$

$F=6.5\times 14.64+6.5\times 9.8$

$F=158.86 N$