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

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

Answer 2

The magnitude of force when time is 4.1s is 124 N

Given information:

mass m=5.00 kg

equation of position $y(t)=(2.80)t+(0.61)^3t^3$

Now the double time derivative of position gives the acceleration:

$a(t)=\frac{d^2y}{dt^2}=\frac{d^2}{dt^2}[2.8t+0.61^3t^3]\\\\a(t)= 3.66t$

Now, at t = 4.10s acceleration is:

$a=3.66\times(4.10)\\\\a=15m/s^2$

Equation of motion:

The equation of motion can be written as:

$F-mg=ma$

where F is the applied upward force and mg is the weight of the crate and ma is the net force acting on the crate.

$F=m(g+a)\\\\F=5(9.8+15)N\\\\F=5\times24.8N\\\\F\approx124N$

So the upward applied force is 124N

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