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Consider an object consisting of two balls connected by a spring, whose stiffness is460 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant the spring is stretched 0.37 m, and the two balls at the ends of the spring have the following masses and velocities:
⢠1: 8 kg, ⹠4, 11, 0 ⺠m/s


⢠2: 4 kg, â¹ â3, 10, 0 ⺠m/s


(a) For this system, calculate

p with arrowsys = ? kg · m/s

(b) Calculate v with arrowCM ? = m/s


(c) Calculate Ktot ?= J


(d) Calculate Ktrans ? = J


(e) Calculate Krel ? = J

Respuesta :

Answer:

V =  1.6777 m/s , 10.667 , 0 m/s

P = ( 20, 128, 0 ) kg-m/s

[tex]K_{tot}[/tex] = 766 J

[tex]K_{trans}[/tex] =  699.3 J

Krel = 66.7 J

Explanation:

given data

Stiffness K = 460 N/m

Extension e =  0.37 m

solution

we know that here that center of mass is express as

V =  [tex]\frac{m1\times v1 + m2\times v2}{m1+m2}[/tex]     ................1

V1 = [tex]\frac{8\times 4 + 4\times -3}{8+4}[/tex]  

V1 =  1.6777 m/s

V2 =  [tex]\frac{8\times 11 + 4\times 10}{8+4}[/tex]  

V2 = 10.667 m/s

V3 =  [tex]\frac{8\times 0 + 4\times 0}{8+4}[/tex]  

V3 = 0 m/s

and

P system is

P = [tex]M_{tot} \times V_{cm}[/tex]

P = (8+4) × ( 1.6777 , 10.667 , 0 )

P = ( 20, 128, 0 ) kg-m/s

and

[tex]K_{tot}[/tex] = K1 + K2

[tex]K_{tot}[/tex] = 0.5 × m × v1² + 0.5 × m × v2²

[tex]K_{tot}[/tex] = 0.5 × 8 ×  [tex]\sqrt{4^2+11^2+0^2}[/tex] + 0.5 × 4 × [tex]\sqrt{-3^2+10^2+0^2}[/tex]  

[tex]K_{tot}[/tex] = 766 J

and

[tex]K_{trans}[/tex] = 0.5 × [tex]M_{tot}[/tex] × ( V(cm)²)

[tex]K_{trans}[/tex] = 0.5 × (8+4) × [tex](\sqrt{1.667^2+12.667^2})^2[/tex]

[tex]K_{trans}[/tex] =  699.3 J

and  

Krel = [tex]K_{tot}[/tex] - [tex]K_{tot}[/tex]

Krel = 766 J  - 699.3 J

Krel = 66.7 J

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