Question

A 6.00-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/s^3)t^3. What is the magnitude of F when t = 4.00 s?

Answer 1

Answer:

The magnitude of F when t=4 s=146.64 N

Explanation:

We are given that

Mass of crate,m=6.00 kg

Height of crate above its initial position is given by

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

We have to find the magnitude of F when t=4.00 s

Differentiate w.r.t t

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

$\frac{d^2y}{dt^2}=6(0.61)t$

$a(t)=\frac{d^2y}{dt^2}=6(0.61)t m/s$

$a(4)=6(0.61)(4)=14.64 m/s^2$

Now, magnitude of force

$F=m(a+g)=6(14.64+9.8)=146.64N$

Hence, the magnitude of F when t=4 s=146.64 N