Respuesta :
Answer:
6833.33 N
Explanation:
By impulse-momentum theorem, change of momentum (p) is equal to impulse (J):
[tex] J=p_f-p_i[/tex]
With pf the momentum at the woman stops (is zero due that), and pi the momentum of the woman just before hits the net ([tex] p_i=mv[/tex])
[tex] J=-mv[/tex] (1)
with v the velocity, that velocity can be fin using conservation of energy:
[tex] [/tex][tex]K=U[/tex]
with U potential energy, K kinetic energy
[tex] \frac{mv^2}{2}=mgh [/tex]
with m the mass, v the velocity, g gravitational acceleration and h the height the woman falls. Solving for v
[tex]v=\sqrt{2gh}=\sqrt{2(9.81)(10.0)}=14.0\frac{m}{s} [/tex] (2)
Using (2) on (1):
[tex]J=-(60.0)(14.0)=820\frac{kg*m}{s} [/tex]
But impulse is too the product of average force (Favg) and the time (t) the force is applied so:
[tex] F_{avg}*t=J [/tex]
Solving for Favg:
[tex] F_{avg}=\frac{J}{t}=\frac{820}{0.120}=-6833.33 N[/tex]
The negative sign indicates the force is opposite to the girls fall direction.
Impulse is the product of force and time. The force exerted by the net on women is 6833.33 N in the opposite direction.
Impulse:
From the impulse-momentum theorem, change in momentum is impulse,
[tex]J = p_f-p_i[/tex]
Since, final momentum is zero,
[tex]J = - p_i\\\\J= - mv[/tex].....................1
Since kinetic energy = potential energy
So,
[tex]\dfrac 12 mv^2 = mgh\\\\v = \sqrt{gh}[/tex]
Put the values,
[tex]v= \sqrt {2 \times 9.8\times 10}\\\\v = 14 m/s[/tex]
Put these values in equation 1,
[tex]J = -60 \times 14\\\\J = - 820\rm \ N/s[/tex]
Since, the impulse is the product of times and force,
[tex]J = F \times s\\\\F = \dfrac{ -820} {0.12}\\\\F = -6833.33 \rm\ N[/tex]
Therefore, the force exerted by the net on women is 6833.33 N in the opposite direction.
Learn more about Impulse:
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