Answer:
The final speed of the masses is 8.75 m/s.
Explanation:
Given that,
Mass = 8.9 kg
Other mass = 2.9 kg
Distance = 7.7 m
Let mass [tex]M_{s}[/tex] start at its 0 Potential energy position.
Let mass [tex]M_{l}[/tex] start with potential energy from its position 4.6 m over final position.
We need to calculate the final speed of the masses
Using conservation of energy
[tex]K.E_{s}+P.E_{s}+K.E_{l}+P.E_{l}=K.E'_{s}+P.E'_{s}+K.E'_{l}+P.E'_{l}[/tex]
[tex]\dfrac{1}{2}m_{s}v^2+m_{s}gh+\dfrac{1}{2}m_{l}v^2+m_{l}gh=\dfrac{1}{2}m_{s}v^2+m_{s}gh+\dfrac{1}{2}m_{l}v^2+m_{l}gh[/tex]
Put the value in the equation
[tex]0+0+0+8.9\times9.8\times7.7=\dfrac{1}{2}m_{s}v^2+2.9\times 9.8\times7.7+\dfrac{1}{2}m_{l}v^2+0[/tex]
[tex]8.9\times9.8\times7.7-2.9\times9.8\times7.7=\dfrac{1}{2}v^2(8.9+2.9)[/tex]
[tex]\dfrac{1}{2}v^2\times11.8=452.76[/tex]
[tex]v^2=\dfrac{452.76\times2}{11.8}[/tex]
[tex]v=8.75\ m/s[/tex]
Hence, The final speed of the masses is 8.75 m/s.
The final speed of the masses after the heavier mass has fallen is 8.76 m/s.
Given the following data:
To determine the final speed of the masses after the heavier mass has fallen 7.7 meters, we would apply the law of conservation of energy:
In accordance with the law of conservation of energy, the potential energy possessed by the masses at the beginning is equal to the kinetic energy possessed by the masses after falling:
[tex]P.E = K.E\\\\m_Agh = m_Bgh+\frac{1}{2} m_Av^2+\frac{1}{2} m_Bv^2\\\\8.9 \times 9.8 \times 7.7=2.9 \times 9.8 \times 7.7+ \frac{1}{2}\times (8.9+2.9)v^2\\\\671.594=218.834+5.9v^2\\\\5.9v^2=671.594-218.834\\\\5.9v^2=452.76\\\\v=\sqrt{\frac{452.76}{5.9} } \\\\v=\sqrt{76.74}[/tex]
Final speed, v = 8.76 m/s.
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