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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.
a. What is the magnitude of the force F when 4.10s ?b. is the magnitude's unit N but the system doesn't accept it?

Respuesta :

Answer

F = 124 N

Explanation:

given,

mass, m = 5 Kg

time, t = 4.1 s

displacement = y(t)=(2.80 m/s)t +(0.61 m/s³)t³

velocity

[tex]\dfrac{dy(t)}{dt}=2.80 + 1.83 t^2[/tex]

[tex]v=2.80 + 1.83 t^2[/tex]

again differentiating to get the equation of acceleration

[tex]\dfrac{dv}{dt}= 3.66 t[/tex]

[tex]a= 3.66 t[/tex]

force at time t = 4.10 s

F = m a

F = 5 x 3.66 x 4.1

F = 75 N

the net force when crate is moving upward

F = Mg + Ma

F = 5 x 9.8 + 75

F = 124 N

the magnitude of force is equal to 124 N

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