91Ó°ÊÓ

Newton's Second Law is fundamental in understanding how forces affect motion. It states that the acceleration of an object depends on two factors: the net force acting on the object and the object's mass. Mathematically, this is expressed as:\[ F_{\text{net}} = m \times a \]Where:

• \( F_{\text{net}} \) is the net force applied to the object,
• \( m \) is the mass of the object,
• \( a \) is the acceleration produced.

Understanding this law allows us to predict how an object's velocity changes when subjected to a certain force. In our exercise, the runner's foot experiences a large force on each step, which results in significant acceleration despite the runner's weight.

The force ratio is an important concept in measuring how much greater the force is compared to a reference point, in this case, the runner's own weight. It helps us understand the magnitude of forces involved during motion. In real-world scenarios like running, this ratio can show how much stress is put on the body. Calculating the force ratio involves the division of the exerted force by the person's body weight:\[\text{Ratio} = \frac{\text{Force exerted on foot}}{\text{Body Weight}}\]From the exercise:

• The force exerted on the foot is \( 2300 \, \text{N} \).
• The body weight of the person is \( 656.37 \, \text{N} \).

Substituting these values, the force ratio is approximately 3.5. This ratio indicates how intense the force on the foot is compared to the gravitational pull on the runner's body weight.

Once we know the net force acting on the runner, we can use it to find the acceleration. In this context, acceleration quantifies how fast the velocity of the runner is changing. To calculate acceleration, we revisit Newton's Second Law:\[F_{\text{net}} = m \times a\]We rearrange the formula to solve for acceleration:\[ a = \frac{F_{\text{net}}}{m}\]For our particular problem, the net force is the difference between the force exerted by the ground and the person's weight. Plugging the values into the formula gives:

• Net force \( F_{\text{net}} = 1643.63 \, \text{N} \)
• Mass \( m = 67 \, \text{kg} \)

Thus, the acceleration is approximately \( 24.53 \, \text{m/s}^2 \), meaning the force from the ground generates a significant boost to the runner's speed.

The change in velocity gives insights into how quickly an object speeds up or slows down over time. In kinematics, the relationship between acceleration and velocity change is straightforward. If we know the time over which the acceleration occurs, we can find the velocity change using:\[\Delta v = a \times t\]Here:

• \( a \) is the acceleration (\( 24.53 \, \text{m/s}^2 \)).
• \( t \) is time duration (\( 10.0 \, \text{ms} = 0.01 \, \text{s} \)).

Multiplying the acceleration by the time period gives us the change in the vertical component of the velocity:\[\Delta v = 0.2453 \, \text{m/s}\]This means the runner's vertical speed changes by this amount during the short contact time, showing a rapid pace modulation in each step.