Problem 47 - University Physics with Modern Physics

Question

A 75.0-kg man steps off a platform 3.10 m above the ground. He keeps his legs straight as he falls, but his knees begin to bend at the moment his feet touch the ground; treated as a particle, he moves an additional 0.60 m before coming to rest. (a) What is his speed at the instant his feet touch the ground? (b) If we treat the man as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant? (c) Draw his freebody diagram. In terms of the forces on the diagram, what is the net force on him? Use Newton's laws and the results of part (b) to calculate the average force his feet exert on the ground while he slows down. Express this force both in newtons and as a multiple of his weight.

Step-by-Step Solution

Answer

(a) 7.80 m/s (b) 50.7 m/s² upward (c) 4538.25 N (approx. 6.17 times his weight)

Step 1: Calculate Initial Speed Before Impact

To find the speed of the man as he touches the ground, we will use the energy conservation principle (or alternatively the kinematic equation for free-fall if preferred). When the man steps off the platform, he falls under the influence of gravity. His potential energy at the top is converted into kinetic energy at the bottom. The potential energy lost is equal to the kinetic energy gained.The potential energy at the top is given by \(PE = mgh\), where \(m\) is the mass (75.0 kg), \(g\) is the gravity (9.81 m/s²), and \(h\) is the height (3.10 m).The kinetic energy at the bottom just before impact is \(KE = \frac{1}{2}mv^2\).Set the potential energy equal to the kinetic energy:\[ mgh = \frac{1}{2} mv^2 \]Cancel out \(m\) and solve for \(v\):\[ gh = \frac{1}{2} v^2 \]\[ v = \sqrt{2gh} \]\(v = \sqrt{2 \cdot 9.81 \cdot 3.10} = \sqrt{60.882} \approx 7.80 \text{ m/s}\).

Step 2: Calculate Acceleration After Impact

Once the man's feet touch the ground, he comes to a stop after moving 0.60 m. We can use the kinematic equation \(v_f^2 = v_i^2 + 2a \cdot d\) to find the acceleration, where \(v_f = 0\) m/s is the final velocity, \(v_i = 7.80\) m/s is the initial velocity, \(a\) is the acceleration, and \(d = 0.60\) m is the distance.Rearrange the equation to solve for \(a\):\[ 0 = 7.80^2 + 2a \cdot 0.60 \]\[ a = \frac{-7.80^2}{2 \cdot 0.60} \]\[ a \approx \frac{-60.84}{1.20} \]\[ a \approx -50.7 \text{ m/s}^2 \]The acceleration is \(-50.7 \text{ m/s}^2\). The negative sign indicates the acceleration is directed upward, opposite the motion.

Step 3: Draw Free Body Diagram

The free body diagram of the man just after touching the ground involves two main forces acting on him: the gravitational force (weight) acting downward \(F_g = mg = 75.0 \cdot 9.81\) N, and the normal force \(F_n\) exerted by the ground acting upwards. Since the man comes to rest, the normal force must also include the force exerted by the ground to stop him.

Step 4: Calculate Net Force and Normal Force

Use Newton's second law \(F_{net} = ma\) to find the net force:\[ F_{net} = 75.0 \cdot (-50.7) \approx -3802.5 \text{ N} \]Since \(F_{net} = F_n - F_g\), we find:\[ F_n = F_g + |F_{net}| \]\[ F_g = 75.0 \cdot 9.81 \approx 735.75 \text{ N} \]\[ F_n = 735.75 + 3802.5 \approx 4538.25 \text{ N} \]The average force exerted by his feet on the ground is approximately 4538.25 N. In terms of his weight, this force is:\[ \text{Force multiple} = \frac{4538.25}{735.75}\]\[ \approx 6.17 \times \text{his weight} \]

Step 5: Summary and Final Results

The speed at which the man's feet touch the ground is approximately 7.80 m/s. After contacting the ground, his acceleration is approximately -50.7 m/s² upward. The average force exerted on the ground is approximately 4538.25 N, equivalently about 6.17 times his body weight.

Key Concepts

Kinematics

Kinematics is a branch of physics that describes the motion of objects without considering the forces causing the motion. It involves parameters such as displacement, velocity, and acceleration. In this problem, we focus on the man's velocity as he falls from the platform. Using the kinematic equations, we derive his speed at impact — just before he hits the ground.

The equation \[ v = \sqrt{2gh} \] where \( v \) is the final velocity, \( g \) is the acceleration due to gravity, and \( h \) is the height fallen, helps us find his speed. It shows that as height increases, the speed at impact also increases if he falls freely under gravity.

Energy Conservation

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the man's potential energy when he is on the platform is converted to kinetic energy as he falls.

Potential energy at height is calculated using \[ PE = mgh \] where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height.

Just before touching the ground, all this potential energy has been converted into kinetic energy, expressed by:\[ KE = \frac{1}{2}mv^2 \]

Equating these two expressions (since \( PE \) = \( KE \) at the point before impact) allows us to solve for the speed at the moment of impact.

Newton's Laws of Motion

Newton's Laws of Motion describe the relationship between a body and the forces acting upon it. In this problem, the second law, which states that the acceleration of a body is directly proportional to the net force acting upon it and inversely proportional to its mass, is used.

Once the man hits the ground, he decelerates until he stops. This deceleration, expressed as a negative acceleration, tells us the net force acting against his motion:\[ F_{net} = ma \]

His weight acts downward, while the ground exerts an upward force (normal force) to stop him. Using the calculated acceleration, we can solve for the net force, helping us understand how his motion ceases due to this opposing force.

Free Body Diagram

A free body diagram is a graphical illustration used to visualize the forces acting on an object. When the man's feet touch the ground, the forces involved include his weight (a downward force due to gravity) and the normal force (an upward force exerted by the ground).

Drawing these forces, we see:

• The gravitational force \( F_g = mg \)
• The normal force \( F_n \), which is greater than the gravitational force because it must decelerate the man to rest

The net force is the result of these opposing forces and can be calculated using the difference. Knowing the forces and using Newton's second law, the average force exerted by his feet on the ground is determined, which is crucial for understanding the impact forces during collisions.