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Chapter 5: Problem 69

\(\mathrm{a}\) (a) A 75-kg man steps out a window and falls (from rest) \(1.0 \mathrm{~m}\) to a sidewalk. What is his speed just before his feet strike the pavement? (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately \(0.50-\mathrm{cm}\) give in the pads of his feet. Calculate the average force exerted on him by the ground in this situation. This average force is sufficient to cause damage to cartilage in the joints or to break bones.

Short Answer

Expert verified
(a) The man's final speed just before he hits the ground is approximately \(4.43 \, m/s\). \n (b) The average force exerted on the man by the ground if he lands with no cushion other than the pads on his feet is approximately \(58,800 \, N\). This is a huge force and it indicates a huge possibility of causing damage to the bones or to the cartilage in the joints.

Step by step solution

01

Calculate the final speed of the man

Since the man starts from rest and falls under the acceleration due to gravity, we can use the equation for motion under constant acceleration, \(v^2 = u^2 + 2g \cdot h\), where \(v\) is the final speed, \(u\) is the initial speed (0 in this case), \(g\) is the acceleration due to gravity (approximately 9.8 m/s^2), and \(h\) is the height from which the man falls (1.0 m). Solving the equation gives \(v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 1.0} \approx 4.43 \, m/s\).
02

Calculate the average force exerted on the man

The force exerted on the man by the ground as he lands can be calculated using the equation for force \(F = \Delta p/ \Delta t\), where \(\Delta p\) is the change in momentum and \(\Delta t\) is the change in time. The change in momentum is the final momentum minus the initial momentum, or \(m \cdot v - mv_0 = m \cdot v\), since the initial velocity \(v_0\) is 0. The change in time can be approximated using the equation \(\Delta t = \Delta h / v\), where \(\Delta h\) is the give in the pads of his feet (0.50 cm = 0.005 m). Substituting these values results into the equation gives \(F = m \cdot v / (\Delta h / v) = m \cdot v^2 / \Delta h\). Substituting the given values results into \(F = 75 \cdot (4.43)^2 / 0.005 = 58,800 \, N\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Fall
When an object is in free fall, it moves solely under the influence of gravity. This means there are no other forces acting on it, like air resistance. Free fall occurs in a vacuum, where all objects fall at the same rate, regardless of their mass.

In our example, a 75-kg man steps out of a window. As he falls, the only force acting on him is gravity. The acceleration due to gravity is a constant 9.8 m/s虏 near the Earth's surface. Because he starts from rest, his initial velocity is 0 m/s. From the moment he begins his fall, his speed increases by 9.8 m/s every second due to gravitational acceleration.

Understanding free fall is crucial for calculating the speed just before impact. Knowing how gravity works in these situations helps predict how fast an object will travel over a certain distance when only gravity is affecting it.
Equations of Motion
In physics, the equations of motion describe how the velocity and position of an object change over time. These equations are especially useful for calculating various parameters of moving objects under uniform acceleration, like the acceleration due to gravity in free fall.

In the exercise, we used one of these key motion equations: the one that relates velocity, acceleration, and height, which is \( v^2 = u^2 + 2g \cdot h \). Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( h \) is the height fallen.

This equation is particularly handy when you know the initial velocity, the fall acceleration, and the distance fallen 鈥� perfect conditions for free fall problems. Knowing the final velocity just before hitting the ground helps in calculating further impacts, like the force experienced upon collision.
Force Calculation
Force is the interaction that causes an object to accelerate. To find the average force exerted by the ground on the man as he hits, we need to look at the change in momentum over a very short time during the impact.

Momentum (\( p \)) is the product of mass and velocity (\( p = mv \)). When the man hits the ground, his velocity changes from 4.43 m/s to 0 m/s. This change in momentum happens quickly as he slows to a stop.

The force exerted can be calculated using Newton's second law in the form \( F = \Delta p / \Delta t \). Because the time \( \Delta t \) is very short, it's important to accurately determine it to find the force. From the solution, \( \Delta t \) was estimated based on the distance his feet cushioned him, giving us a final force of approximately 58,800 N.

Understanding these calculations is key to grasping why impacts can be so damaging, illustrating how powerful forces act over very short times during an impact.

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Most popular questions from this chapter

A 70 -kg diver steps off a \(10-\mathrm{m}\) tower and drops from rest straight down into the water. If he comes to rest \(5.0 \mathrm{~m}\) beneath the surface, determine the average resistive force exerted on him by the water.

A loaded ore car has a mass of \(950 \mathrm{~kg}\) and rolls on rails with negligible friction. It starts from restand is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at \(30.0^{\circ}\) above the horizontal. The car accelerates uniformly to a speed of \(2.20 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\) and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length \(1250 \mathrm{~m} ?\)

(a) A child slides down a water slide at an amusement park from an initial height \(h\). The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-of-energy equation be modified? (c) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is \(12.0 \mathrm{~m}\).

A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Active Figure \(5.20 .\) The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through cquilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

A \(60.0\) -kg athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0 .\) What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.
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