Change in momentum: Child jumping from a swing on a playground

[hendrix7]

Homework Statement

A child is on a playground swing. The playground has a rubber safety
surface.
The child, of mass 32 kg, jumped from the swing.
When the child reached the ground she took 180 milliseconds to slow down and stop.
During this time an average force of 800 N was exerted on her by the ground.
Calculate the velocity of the child when she first touched the ground.

Relevant Equations

Force = change in momentum / time

800 - (32 x 9.8) = 32v/0.18 where v = velocity
this gives me v = 2.736 m/s
The answer given, however, is 800 = 32v/0.18, i.e. v = 4.5 m/s
The difference, of course, is the weight of the child. I don't understand why this is not allowed for in the net force acting on the child. Can someone put me straight here?

 

[gneill]

Homework Statement: A child is on a playground swing. The playground has a rubber safety
surface.
The child, of mass 32 kg, jumped from the swing.
When the child reached the ground she took 180 milliseconds to slow down and stop.
During this time an average force of 800 N was exerted on her by the ground.
Calculate the velocity of the child when she first touched the ground.

I think that the force due to gravity was included in the average force exerted upon her from the ground.

 

[TSny]

I interpret the force "exerted on her by the ground" as the normal force from the ground, which would not include the child's weight. So, I like @hendrix7 's solution (2.7 m/s). Just my 2 cents.

 

[kuruman]

Let's see what Sir Isaac has to say.
$F_{\text{net}}=N-mg=ma$
We average the forces over a time interval $\Delta t$. The time-average of $mg$ is $mg$.
$\bar {F}_{\text{net}}=\bar N-mg.$
$\bar {F}_{\text{net}}=m\bar a=m\dfrac{\Delta v}{\Delta t}.$
Thus, $m\dfrac{\Delta v}{\Delta t}=\bar N-mg.$
The force exerted on the child by the ground is the normal force $N$. We are told that the average force exerted on her by the ground during this time interval is 800 N. Thus, $\bar N = 800~$N. Substituting, $$32~(\text{kg}) \frac{0-v_0~(\text{m/s})}{0.18~(\text{s})}=800~(\text{N})-32~(\text{kg})*9.8~(\text{m/s}^2)\implies v_0=-2.7~(\text{m/s}).$$The negative sign says that $v_0$ is "down" in the same direction as the acceleration of gravity.

I think that this solution is transparent and sound. The answer given is incorrect.

 

[haruspex]

The force exerted on the child by the ground is the normal force N.

Unless ground includes the whole Earth, as @gneill suggests. But I agree, the given answer is a blunder.

 

[kuruman]

Unless ground includes the whole Earth, as @gneill suggests. But I agree, the given answer is a blunder.

If the idea were to include the Earth, then the given information should have been something like "During this time an average net force of 800 N was required to stop her." Probably the author didn't think there is a difference between "net" and "ground" since, after all, it's the ground that stops the girl.

 

[hendrix7]

If the idea were to include the Earth, then the given information should have been something like "During this time an average net force of 800 N was required to stop her." Probably the author didn't think there is a difference between "net" and "ground" since, after all, it's the ground that stops the girl.

Thanks, kuruman, that makes sense to me, and thanks to everyone else who replied.