A 45 kg woman jumps from a ledge that is 1.2 m above concrete.

Question

(a) What is her speed just before her feet strike the pavement?

(b) The only cushion for her feet is the 0.50 cm in the pads of her feet. So, she displaces only 0.50 cm during the landing. Calculate the average force the ground exerts on her in this situation.

Answer 1

Hi Charlotte A.!

Part One:

There are several approaches to answer the first part of this question. For instance, we could consider the conservation of energy:

E_o + W_nc = E_f

Because we haven't any non-conservative forces acting on our system, we can drop that out of our equation:

E_o = E_f

Now, we must consider which forces we have acting at the initial and final positions. Because the woman jumps from a ledge, we know that we have a non-zero height. Moreover, because she exists within a gravitational field, we know that her non-zero height confers gravitational potential energy.

Additionally, if we take the ground to be a height of zero, then we can say she no longer has gravitational potential energy at this position. Instead, the conservation of energy tells us that all of her initial gravitational potential energy has been converted in kinetic energy.

We are now ready to tease out our equation further:

Ug_o = KE_f

mgh_o = (1/2)mv_f²

Notice that her mass cancels because it appears in all of our terms:

gh_o = (1/2)v_f²

v_f = √2gh_o

<=> √(2)(9.8 m*s^-1)(1.2 m)

v_f = 4.8 m*s^-1

Note because the question asked for speed, we would take the magnitude only and not consider the direction.

As we discussed above, this is not the only solution. Consider, we could have simply used a kinematics equation and obtained the same conclusion:

v_f² = v_o² + 2aΔy

v_f² = 2gΔy

v_f = √2gΔy

Note that both Δy and g are negative, so the value under the square root is not imaginary.

Part Two:

Starting with a free-body diagram would be wise here so can visualize everything that is acting on our system (the woman). During its construction, you would notice that we have the force due to gravity pointing downward, and the force due to the concrete pointing upward. We are ready to formulate a Newton's 2nd Law equation:

ΣF = F_concrete - F_g = ma

While we have enough information to calculate the force due to gravity, we need to consider whether we can say the same for the net force. In this scenario, we know only the mass. We do not know the acceleration. In this case, the acceleration will not be equal to 9.8 m*s^-2 because the woman is not in free-fall. She is involved in a collision with the concrete in this part, so she is not simply falling freely from the sky as she was in the first part. That means we have two unknowns, the force due to the concrete and the acceleration, but we have only one equation. We need another equation. Specifically, we need an equation that describes the net acceleration.

What else do we know?

1. We know that she travels 0.5 cm
2. We know that her initial velocity is equal to the final velocity we found in the first part
3. We know her final velocity is 0 m*s^-1 because she comes to a stop

This information suggests we use a kinematic equation to find her acceleration.

v_f² = v_o² + 2aΔy

=> v_f = 0 m*s^-1

0 = v_o² + 2aΔy

0 - v_o² = 2aΔy => a = -v_o²/2Δy

Note that both the square of the initial velocity and displacement are negative values, and so the negatives will cancel to yield a positive acceleration--in other words, an acceleration pointing upward in the positive y-direction.

a = v_o²/2Δy

Now, we are ready to solve our Newton's 2nd Law equation.

ΣF = F_concrete - F_g = ma

F_concrete = ma + F_g = m(a + g)

Let us substitute our expression for the acceleration into our equation:

F_concrete = m(a + g)

=> a = v_o²/2Δy

F_concrete = m[(v_o²/2Δy) + g]

<=> (45kg)[(4.8 m*s^-1)²/(2*0.05m) + 9.8m*s^-2]

F_concrete = 10809 N

Hope that helps!

Cheers