Question

A 5.00 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)=(2.80 m/s)t +(0.61 m/s^3)t^3.1. What is the magnitude of the force F when 4.10s ?2. is the magnitude's unit N but the system doesn't accept it?

Answer 1

Answer:

F=124.03 N

Explanation:

Given data

mass m=5.00 kg

y(t)=(2.80 m/s)t+(0.61m/s³)t³

To find

Force F at t=4.10 s

Solution

As

$y(t)=(2.80m/s)t+(0.61m/s^{3} )t^{3}$

derivative with respect to time we get

$\frac{dy(t)}{dt} =v(t)=2.80+1.83t^{2}$

again derivative with respect to time to get acceleration

$\frac{dv(t)}{dt}=a(t)=3.66t$

at t=4.10 s then a(t) is

a=3.66(4.10)

a=15.006 m/s²

Now to use Newton law to find force in y direction

$F-mg=ma\\F=mg+ma\\F=(5.00kg)(9.8m/s^{2} )+(5.00kg)(15.006m/s^{2} )\\F=49N+75.03N\\F=124.03N$