(a) The velocity is "6.2 m/s".
(b) The average force is "800.83 N".
According to the question,
Force,
Deflection,
= [tex]1.2\times 10^{-2} \ m[/tex]
As we know,
The work done,
→ [tex]W = F\times d[/tex]
[tex]= 800\times 1.2\times 10^{-2}[/tex]
[tex]= 9.6 \ J[/tex]
(a)
Given:
Mass of hand,
Now,
→ [tex]\frac{1}{2} mv^2 = 9.6 \ J[/tex]
[tex]v = \sqrt{\frac{2\times 9.6}{0.50} }[/tex]
[tex]= 6.2 \ m/s[/tex]
(b)
→ [tex]v^2 = u^2 +2ax[/tex]
→ [tex]a= \frac{v^2}{2x}[/tex]
[tex]= \frac{(6.2)^2}{2\times 1.2\times 10^{-2}}[/tex]
[tex]= 1601.67 \ m/s^2[/tex]
hence,
The average force will be:
→ [tex]F_{avg} = m\times a[/tex]
[tex]= 0.50\times 1601.67[/tex]
[tex]= 800.83 \ N[/tex]
Thus the above answers are correct.
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