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

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

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

$F=158.86 N$

Answer 2

The magnitude of the force acting on the crate is 95.16N

Equation

An equation is an expression used to show the relationship between two or more numbers and variables.

Given the equation for height as:

y(t) = 2.8t + 0.61t³

The velocity (v) is:

v(t) = y'(t) = 2.8 + 1.83t²

The acceleration (a) is:

a(t) = v'(t) = 3.66t

At 4 seconds:

a = 3.66(4) = 14.64 m/s²

Force (F) = mass * acceleration = 6.5 kg * 14.64 m/s² = 95.16N

The magnitude of the force acting on the crate is 95.16N

Find out more on Equation at: https://brainly.com/question/13763238