Problem 56
Question
Jumping to the ground. A 75.0 \(\mathrm{kg}\) man steps off a platform 3.10 \(\mathrm{m}\) above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.60 \(\mathrm{m}\) before coming to rest. (a) What is his speed at the instant his feet touch the ground? (b) Treating him as a particle, what are the magnitude and direction of his acceleration as he slows down if the acceleration is constant? (c) Draw a free-body diagram of this man as he is slowing down. (d) Use Newton's laws and the results of part (b) to calculate the force the ground exerts on him while he is slowing down. Express this force in newtons and also as a multiple of the man's weight. (e) What are the magnitude and direction of the reaction force to the force you found in part (c)?
Step-by-Step Solution
Verified Answer
(a) 7.80 m/s; (b) 50.7 m/s² upward; (c) Diagram as described; (d) 4537.5 N, 6.17 times weight; (e) Reaction force is 4537.5 N downward.
1Step 1: Calculate speed when feet touch the ground
We use the kinematic equation to find the speed when the man's feet touch the ground. The equation for velocity is \( v^2 = u^2 + 2as \), where \( u = 0 \text{ m/s} \) (initial velocity), \( a = 9.8 \text{ m/s}^2 \) (acceleration due to gravity), and \( s = 3.10 \text{ m} \) (distance fallen). Thus:\[v^2 = 0 + 2 imes 9.8 imes 3.10 = 60.76\]Solving for \( v \), we get \( v = \sqrt{60.76} \approx 7.80 \text{ m/s} \).
2Step 2: Determine acceleration while slowing down
We now find the acceleration as he comes to rest over the additional distance (0.60 m). Applying the kinematic equation \( v^2 = u^2 + 2as \) again, where \( v = 0 \text{ m/s} \), \( u = 7.80 \text{ m/s} \), and \( s = 0.60 \text{ m} \):\[0 = (7.80)^2 + 2a(0.60)\]This simplifies to:\[a = \frac{-60.84}{1.20} = -50.7 \text{ m/s}^2\]Thus, the acceleration is \(-50.7 \text{ m/s}^2\), upward.
3Step 3: Create a free-body diagram
In the free-body diagram, depict the man during deceleration. Draw two main forces:1. Gravitational force (\( F_g = mg = 75.0 \cdot 9.8 \text{ N} \)) acting downwards.2. Normal force exerted by the ground (\( F_n \)) acting upwards as the man comes to rest.
4Step 4: Calculate force exerted by ground
We apply Newton’s second law, \( F = ma \), using the acceleration found in Step 2. Considering the total force exerted by the ground:\[F_n - F_g = ma\]\[F_n - 75.0 \times 9.8 = 75.0 \times 50.7\]\[F_n = 75.0 \times 9.8 + 75.0 \times 50.7 = 75.0 (9.8 + 50.7) = 75.0 \times 60.5 = 4537.5 \text{ N}\]So, the force is \( 4537.5 \text{ N} \). As a multiple of the man's weight: \( \frac{4537.5}{735} \approx 6.17 \).
5Step 5: Determine reaction force
According to Newton's third law, the ground exerts a force of \( 4537.5 \text{ N} \) upward on the man, so the reaction force is \( 4537.5 \text{ N} \) exerted by the man downward on the ground.
Key Concepts
Newton's Laws of MotionFree-Body DiagramAccelerationForce Calculation
Newton's Laws of Motion
Newton's Laws of Motion are fundamental to understanding the dynamics of physical systems. These laws describe the relationship between the motion of an object and the forces acting on it. Let's focus on each one:
1. **First Law** (Law of Inertia): An object at rest stays at rest, and an object in motion continues in motion with the same speed and direction unless acted upon by a net external force. In our exercise, the man remains at rest on the platform until gravity acts, providing an external force that causes his descent.
2. **Second Law** (Law of Acceleration): This law is best captured in the equation: \( F = ma \), where \( F \) is the net force applied, \( m \) is the mass, and \( a \) is the acceleration. In this exercise, the man's acceleration when hitting the ground and during deceleration is a key focus. We calculate the force with this principle.
3. **Third Law** (Action and Reaction): For every action, there is an equal and opposite reaction. The man's force exerted on the ground upon impact is matched by an equal force exerted by the ground on the man, described in part (e) of the original exercise.
1. **First Law** (Law of Inertia): An object at rest stays at rest, and an object in motion continues in motion with the same speed and direction unless acted upon by a net external force. In our exercise, the man remains at rest on the platform until gravity acts, providing an external force that causes his descent.
2. **Second Law** (Law of Acceleration): This law is best captured in the equation: \( F = ma \), where \( F \) is the net force applied, \( m \) is the mass, and \( a \) is the acceleration. In this exercise, the man's acceleration when hitting the ground and during deceleration is a key focus. We calculate the force with this principle.
3. **Third Law** (Action and Reaction): For every action, there is an equal and opposite reaction. The man's force exerted on the ground upon impact is matched by an equal force exerted by the ground on the man, described in part (e) of the original exercise.
Free-Body Diagram
A free-body diagram is a crucial tool in physics to visualize the forces acting on an object. It helps us apply Newton's laws effectively.
When the man hits the ground and comes to rest, we identify two critical forces in the free-body diagram:
These forces are crucial for analyzing the motion and deceleration in the steps following the man's impact with the ground.
When the man hits the ground and comes to rest, we identify two critical forces in the free-body diagram:
- **Gravitational Force** \((F_g)\): The force due to gravity that pulls the man downward. It is calculated as \( F_g = mg \), where \( m \) is the man's mass (75.0 kg) and \( g = 9.8 \) m/s² is the acceleration due to gravity.
- **Normal Force** \((F_n)\): This upward force is exerted by the ground. It acts perpendicular to the surface, counteracting gravity to bring the man to rest.
These forces are crucial for analyzing the motion and deceleration in the steps following the man's impact with the ground.
Acceleration
Acceleration is the rate of change of velocity of an object. It is a vector quantity, meaning it has both magnitude and direction.
In the exercise at hand, two key accelerations are highlighted:
- **During Descend**: Under the influence of gravity alone, the man's acceleration is \( 9.8 \text{ m/s}^2 \) downward, consistent as he falls freely from the platform.
- **Deceleration Upon Impact**: As the man makes contact with the ground and his knees begin to bend, a sizable opposing acceleration occurs. A calculated acceleration of \(-50.7 \text{ m/s}^2\) upward brings the man to rest over an additional distance of 0.60 m. This negative sign indicates deceleration, demonstrating the force's reversal relative to the initial motion.
In the exercise at hand, two key accelerations are highlighted:
- **During Descend**: Under the influence of gravity alone, the man's acceleration is \( 9.8 \text{ m/s}^2 \) downward, consistent as he falls freely from the platform.
- **Deceleration Upon Impact**: As the man makes contact with the ground and his knees begin to bend, a sizable opposing acceleration occurs. A calculated acceleration of \(-50.7 \text{ m/s}^2\) upward brings the man to rest over an additional distance of 0.60 m. This negative sign indicates deceleration, demonstrating the force's reversal relative to the initial motion.
Force Calculation
The calculation of forces provides insights into the interactions between objects. Newton's Second Law, \( F = ma \), facilitates these computations.
Using the information from the acceleration phase when the man's knees bend, we calculate the force exerted by the ground. The man's mass is 75.0 kg and the upward acceleration is \(-50.7 \text{ m/s}^2\):
First, calculate the net force required to produce this deceleration: \( F = 75.0 \times 50.7 = 3802.5 \text{ N} \).
Add this to the gravitational force to find the ground's total force of \( 4537.5 \text{ N} \). This calculation reveals that the force exerted by the ground is \( 6.17 \) times the man's weight, illustrating the substantial effect of the abrupt stop.
Understanding the systematic approach to calculating these forces emphasizes the urge of safety measures in potential high-fall scenarios.
Using the information from the acceleration phase when the man's knees bend, we calculate the force exerted by the ground. The man's mass is 75.0 kg and the upward acceleration is \(-50.7 \text{ m/s}^2\):
First, calculate the net force required to produce this deceleration: \( F = 75.0 \times 50.7 = 3802.5 \text{ N} \).
Add this to the gravitational force to find the ground's total force of \( 4537.5 \text{ N} \). This calculation reveals that the force exerted by the ground is \( 6.17 \) times the man's weight, illustrating the substantial effect of the abrupt stop.
Understanding the systematic approach to calculating these forces emphasizes the urge of safety measures in potential high-fall scenarios.
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