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Neko [114]
3 years ago
6

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 en

d 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?
Physics
1 answer:
masya89 [10]3 years ago
8 0

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

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