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a 25.0kg child jumps to the ground from a structure 1.00m high and comes to rest 0.500s after first contact with the ground. what average force is applied by the child in coming to rest

User Joly
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1 Answer

3 votes

Answer:

Approximately
470\; {\rm N}.

(Assuming that
g = 9.81\; {\rm N \cdot kg^(-1)} and that air resistance is negligible.)

Step-by-step explanation:

Under the assumptions, the acceleration of the child would be constantly
a = g = 9.81\; {\rm N \cdot kg^(-1)} = 9.81\; {\rm m\cdot s^(-2)} while the child was in the air.

Apply the SUVAT equation
v^(2) - u^(2) = 2\, a\, x to find the velocity
v of the child right before landing:


\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x}\end{aligned}, where:


  • u = 0\; {\rm m\cdot s^(-1)} is the initial velocity of the child,

  • a = 9.81\; {\rm m\cdot s^(-2)} is the vertical acceleration, and

  • x = 1.00\; {\rm m} is the vertical displacement (change in height.)

The child is at rest
\Delta t = 0.500\; {\rm s} after contact. During that
0.500\; {\rm s}, velocity would have changed by
\Delta v = \sqrt{u^(2) + 2\, a\, x}. Momentum of the child would have changed by
\Delta p = m\, \Delta v, where
m = 25.0\; {\rm kg} is the mass of the child.

Divide this change in momentum by the duration
\Delta t to find the average net force:


\displaystyle F_{\text{net}} = (m\, \Delta v)/(\Delta t).

There are two forces on the child: upward normal force from the ground
F_{\text{normal}} and downward gravitational attraction
m\, g from the Earth. The resultant force on the child points upwards:


F_{\text{net}} = F_{\text{normal}} - m\, g.

Rearrange this equation to find the normal force on the child:


\begin{aligned}F_{\text{normal}} &= F_{\text{net}} + m\, g \\ &= (m\, \Delta v)/(\Delta t) + m\, g \\ &= \frac{m\, \sqrt{2\, a\, x + u^(2)}}{\Delta t} + m\, g\\ &= \frac{(25.0)\, \sqrt{2\, (9.81)\, (1.00) + 0^(2)}}{0.500}\; {\rm N} + (25.0)\, (9.81)\; {\rm N}\\ &\approx 470\; {\rm N}\end{aligned}.

This normal force from the ground on the child is the reaction to the force that the child exerted on the ground. The two forces will have the same magnitude: approximately
470\; {\rm N}. Hence, the child would have exerted an average force of approximately
470\; {\rm N}\! on the ground during that
0.500\; {\rm s}.

User Paul Walczewski
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