Problem 51 - University Physics with Modern Physics

Question

Jumping to the Ground. A 75.0 -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 is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant? (c) Draw his free-body diagram (see Section 4.6 ). 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 in newtons and also 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, 6.17 times his weight.

Step 1: Calculate the Speed at Impact (a)

First, calculate the speed of the man just before he touches the ground. Use the kinematic equation for free fall:\[ v^2 = u^2 + 2 g h \]where \( u = 0 \) (initial speed), \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 3.10 \, \text{m} \) (height).Substituting the values:\[ v^2 = 0 + 2 \times 9.81 \times 3.10 \]\[ v^2 = 60.882 \]\[ v = \sqrt{60.882} \]\[ v \approx 7.80 \, \text{m/s} \]Thus, the speed at the instant his feet touch the ground is approximately 7.80 m/s.

Step 2: Calculate Acceleration During Stopping (b)

Next, calculate the acceleration as he bends his knees and comes to rest over an additional 0.60 m. Use the equation:\[ v^2 = u^2 + 2 a d \]where \( v = 0 \) (final speed), \( u = 7.80 \, \text{m/s} \) (speed at the ground), and \( d = 0.60 \, \text{m} \).Rearrange to solve for \( a \):\[ 0 = (7.80)^2 + 2 a (0.60) \]\[ 0 = 60.84 + 1.2 a \]\[ a = \frac{-60.84}{1.2} \]\[ a \approx -50.7 \, \text{m/s}^2 \]The acceleration (magnitude) is approximately 50.7 m/s² and its direction is upwards (opposite to the motion).

Step 3: Free-Body Diagram and Net Force (c)

Draw the free-body diagram including:- Gravitational force \( F_g = mg = 75.0 \times 9.81 = 735.75 \, \text{N} \) downward.- Normal force \( F_n \) exerted by the ground upward.The net force (\( F_{\text{net}} \)) is upward:\[ F_{\text{net}} = F_n - F_g \]Use Newton's second law to find \( F_n \):\[ F_{\text{net}} = ma \]\[ F_n - 735.75 = 75.0 \times 50.7 \]\[ F_n = 735.75 + 3802.5 \]\[ F_n = 4538.25 \, \text{N} \]The normal force is 4538.25 N.

Step 4: Express Force as a Multiple of Weight

To find the force as a multiple of his weight:Weight \( W = 735.75 \, \text{N} \)Thus, the multiple is:\[ \frac{F_n}{W} = \frac{4538.25}{735.75} \approx 6.17 \]The force his feet exert on the ground is approximately 6.17 times his weight.

Key Concepts

Kinematic Equations

Kinematic equations are essential in physics for describing the motion of objects. In problems involving constant acceleration, such as a man falling under gravity, they become very useful. These equations link three primary motion parameters: velocity, acceleration, and displacement. For example, the equation \( v^2 = u^2 + 2gh \) allows us to calculate the final speed \( v \) of an object... by knowing the initial speed \( u \), the height \( h \), and the acceleration \( g \).

• Initial Velocity \( u \): This is the speed at which an object begins its motion.
• Final Velocity \( v \): This is the speed at the point of interest, often before an object stops or changes motion.
• Acceleration \( g \): In free fall, this is the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \).
• Displacement \( h \): The distance an object has moved from its initial position.

Using these equations helps predict motion outcomes, such as how fast an object will be traveling at any point before hitting the ground.

Free-Body Diagram

A free-body diagram is a key tool in physics for visualizing the forces acting upon an object. It simplifies a physical situation by isolating the object and illustrating all the forces with vector arrows. In the context of our example— where a man is falling— the free-body diagram features two primary forces:

• Gravitational Force: This is always directed downwards, representing the weight of the object: \( F_g = mg \), where \( m \) is mass and \( g \) is gravitational acceleration.
• Normal Force: This arises when the object touches a surface, pushing back upwards with a force \( F_n \).

By analyzing the diagram, one can easily determine the net force acting on an object. Such diagrams are indispensable when setting up problems using Newton's laws of motion.

Newton's Laws

Newton's laws of motion are fundamental in understanding how forces affect the motion of objects. In particular, Newton's second law, \( F = ma \), connects force to motion, stating that the net force acting on an object is the product of its mass \( m \) and acceleration \( a \). Let's delve deeper into how this applies:

• When analyzing a scenario where an object decelerates, such as when our man bends his knees, we use the second law to calculate the acceleration force exerted by his legs.
• Newton's third law tells us that for every action, there is an equal and opposite reaction. Thus, the force his legs exert upon the ground translates to an upward force exerted by the ground.

Understanding Newton's laws helps solve problems that involve motion dynamics, such as calculating the impact force, shown by manipulating these principles correctly.

Acceleration Calculation

Acceleration calculation is crucial when solving physics problems involving motion. The general formula used is \( v^2 = u^2 + 2ad \), where \( d \) represents additional displacement. To find acceleration \( a \), we rearrange the formula: \( a = \frac{v^2 - u^2}{2d} \).

• In our example, the moment his feet touch the ground, the man is moving at a speed of approximately 7.80 m/s.
• As he decelerates over a distance of 0.60 m to reach a complete stop (\( v = 0 \)), the negative acceleration (opposite the direction of motion) is calculated to be approximately \(-50.7 \, \text{m/s}^2\).
• This negative sign indicates that the forces are acting opposite to his movement, effectively slowing him down.

Properly calculating acceleration allows us to derive insights into the forces involved and to design safe systems to manage them.