Respuesta :
Answer:
[tex]a = 8.175 m/s^2[/tex]
Explanation:
The mass is connected with a string and wrapped around the spool
So here we can say that the force on the mass is due to weight and the tension in string so as per the force equation we have
[tex]mg - T = ma[/tex]
now for the spool we can say it is rotated due to the tension force on it
so we can say
[tex]TR = I\alpha[/tex]
[tex]TR = (\frac{1}{2}m_s R^2)(\frac{a}{R})[/tex]
[tex]T = \frac{1}{2}m_s a[/tex]
now we have
[tex]mg - \frac{1}{2}m_s a = ma[/tex]
[tex]mg = a(\frac{1}{2}m_s + m)[/tex]
now the acceleration is given as
[tex]a = \frac{mg}{\frac{1}{2}m_s + m}[/tex]
now plug in values of all masses
[tex]a = \frac{0.250(9.81)}{\frac{1}{2}(0.100) + 0.250}[/tex]
[tex]a = 8.175 m/s^2[/tex]
The Magnitude of the acceleration when a mass descends will be
[tex]a=8.175\dfrac{m}{s^2}[/tex]
What will be the acceleration of the mass?
It is given that
Mass m= 250gm
spoon radius r=3cm
and weight = 100gm
Since the mass is tied by a rope and this rope is wrapped on the spoon
by balancing the force from newtons law
[tex]mg-T=ma[/tex]
[tex]Tr=I\alpha[/tex]
[tex]Tr=\dfrac{1}{2} m_{s}R^{2}(\dfrac{a}{R})[/tex]
[tex]Tr=\dfrac{1}{2}m_{s}a[/tex]
now we have
[tex]mg-\dfrac{1}{2} m_{s}a=ma[/tex]
[tex]mg=a(\frac{1}{2} m_{s}+m)[/tex]
now the acceleration is given as
[tex]a=\dfrac{m_{g}}{\frac{1}{2} m_{s}+m}[/tex]
[tex]a=\dfrac{0.25(0.981)}{\dfrac{1}{2}(0.1)+0.250 } =8.175\frac{m}{s^2}[/tex]
Thus the Magnitude of the acceleration when a mass descends will be
[tex]a=8.175\dfrac{m}{s^2}[/tex]
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