Problem 63 - College Physics

Question

\(\bullet$$\bullet\) Prevention of hip injuries. People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. Let us investigate the worst-case sce- nario, in which a 55 kg person completely loses her footing (such as on icy pavement) and falls a distance of \(1.0 \mathrm{m},\) the dis- tance from her hip to the ground. We shall assume that the per- son's entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person's speed to 1.3 \(\mathrm{m} / \mathrm{s}\) over a distance of 2.0 \(\mathrm{cm} .\) Find the acceleration (assumed to be constant) of this person's hip while she is slow- ing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see if it is likely to cause injury, calcu- late how long it lasts.

Step-by-Step Solution

Answer

(a) 4.43 m/s; (b) -448.62 m/s², -24674.1 N; (c) 0.007 seconds

Step 1: Calculate the Final Speed before Impact

We can find the speed at which the person reaches the ground using the kinematic equation for a free-falling object: \[ v^2 = u^2 + 2gh \]where \( u = 0 \) m/s (initial speed), \( g = 9.8 \) m/s² (acceleration due to gravity), and \( h = 1.0 \) m (fall distance). Plugging in the values, we get:\[ v^2 = 0 + 2 \times 9.8 \times 1.0 = 19.6 \]Solving for \( v \), we find:\[ v = \sqrt{19.6} = 4.43 \text{ m/s} \]

Step 2: Calculate Acceleration during Slowing Down

When the hip pad reduces the speed to 1.3 m/s over 2.0 cm, we use the kinematic equation:\[ v_f^2 = v_i^2 + 2a d \]where \( v_f = 1.3 \) m/s, \( v_i = 4.43 \) m/s, and \( d = 0.02 \) m. Plugging in the values:\[ (1.3)^2 = (4.43)^2 + 2a(0.02) \]\[ 1.69 = 19.6249 + 0.04a \]Rearranging to solve for \( a \):\[ a = \frac{1.69 - 19.6249}{0.04} = -448.62 \text{ m/s}^2 \]

Step 3: Calculate the Force Exerted by the Hip Pad

The force exerted by the hip pad can be calculated using Newton's second law:\[ F = ma \]where \( m = 55 \) kg and \( a = -448.62 \) m/s². Thus:\[ F = 55 \times (-448.62) = -24674.1 \text{ N} \]The negative sign indicates the force is acting in the opposite direction of motion.

Step 4: Calculate Time of Force Application

To find the duration over which this force acts, use the formula:\[ v_f = v_i + at \]Solving for \( t \):\[ 1.3 = 4.43 + (-448.62)t \]\[ -3.13 = -448.62t \]\[ t = \frac{-3.13}{-448.62} = 0.00698 \text{ seconds} \]

Key Concepts

Acceleration Due to Gravity

When objects fall freely towards Earth, they accelerate due to the force of gravity. This acceleration is commonly denoted by the letter \( g \) and has a standard value of approximately \( 9.8 \, \text{m/s}^2 \). This means that, in the absence of other forces, the velocity of a falling object increases by \( 9.8 \, \text{m/s} \) each second of its fall.
This concept is crucial in solving problems involving free fall, like the one in the original exercise where a person falls from a height. The kinematic equation \( v^2 = u^2 + 2gh \) helps us calculate the final speed \( v \) of a person falling from rest (initial speed \( u = 0 \)) over a distance \( h \).
In the exercise, the person falls 1 meter, and using the acceleration due to gravity, we find their speed just before impact to be \( 4.43 \, \text{m/s} \). This showcases how gravity impacts the speed of falling objects.

Force and Motion

Force and motion are interconnected concepts that describe how objects move when acted upon by forces. A fundamental aspect of motion is how it is influenced by acceleration, which can be either an increase or decrease in speed.
When the person hits the ground, their speed is influenced by the force exerted by the hip pad. This scenario involves deceleration, where the hip pad decreases their speed from \( 4.43 \, \text{m/s} \) to \( 1.3 \, \text{m/s} \).
To calculate the acceleration during this slow down phase, we rearrange the kinematic equation \( v_f^2 = v_i^2 + 2ad \). With the given data, the calculated acceleration is \( -448.62 \, \text{m/s}^2 \). The negative sign indicates a reduction in speed. This is a significant deceleration, illustrating how force can alter motion in a very short span of time, as discussed in this exercise.

Newton's Second Law

Newton's second law of motion provides a fundamental principle to calculate force: \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration. This law tells us that the amount of force needed to move an object is directly related to how massive the object is and how much it needs to change its motion.
In the exercise, the hip pad exerts a force to slow the person's fall. Using the calculated deceleration of \( -448.62 \, \text{m/s}^2 \) and a mass of \( 55 \, \text{kg} \), Newton's second law helps us find the force exerted: \( F = 55 \, \text{kg} \times (-448.62 \, \text{m/s}^2) = -24674.1 \, \text{N} \).
The calculation's negative sign indicates the force direction is opposite to the fall, working against the motion to stop it. This large force, applied over a short distance, is typical in scenarios where protective gear, like hip pads, are used to prevent injuries by dissipating energy rapidly.