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Chapter 10: Problem 69

A Woman Leaps Vertically A \(55 \mathrm{~kg}\) woman leaps vertically from a crouching position in which her center of mass is \(40 \mathrm{~cm}\) above the ground. As her feet leave the floor, her center of mass is \(90 \mathrm{~cm}\) above the ground; it rises to \(120 \mathrm{~cm}\) at the top of her leap. (a) As she is pressing down on the ground during the leap, what is the average magnitude of the force on her from the ground? (b) What maximum speed does she attain?

Short Answer

Expert verified
The average force is 862.4 N, and the maximum speed is approximately 2.424 m/s.

Step by step solution

01

- Understanding the Problem

Identify that the woman鈥檚 mass is given as 55 kg and three key heights of her center of mass: crouching position (40 cm), feet leaving the floor (90 cm), and the top of the leap (120 cm).
02

- Convert Heights to Meters

Convert the given heights from centimeters to meters: \[ h_1 = 0.40 \text{ m}, \, h_2 = 0.90 \text{ m}, \, h_3 = 1.20 \text{ m} \]
03

- Calculate the Work Done by the Woman

The work done is equal to the change in potential energy as she goes from 40 cm to 120 cm: \[ W = mgh_3 - mgh_1 = 55 \times 9.8 \times (1.20 - 0.40) = 55 \times 9.8 \times 0.80 \] \[ W = 431.2 \text{ J} \]
04

- Calculate the Distance Over Which the Force Acts

The distance over which the force acts is from the crouching position to when her feet leave the floor: \[ d = h_2 - h_1 = 0.90 - 0.40 = 0.50 \text{ m} \]
05

- Calculate the Average Force

The average force is the work done divided by the distance: \[ F = \frac{W}{d} = \frac{431.2}{0.50} = 862.4 \text{ N} \]
06

- Calculate Maximum Velocity Using Energy Principles

At the top of the leap, all kinetic energy has been converted to potential energy. Using energy conservation from 90 cm to 120 cm: \[ K.E. = P.E. \] \[ \frac{1}{2}mv^2 = mg(h_3 - h_2) \] Rearranging to solve for velocity, we get: \[ v = \sqrt{2g(h_3 - h_2)} = \sqrt{2 \times 9.8 \times (1.20 - 0.90)} = \sqrt{2 \times 9.8 \times 0.30} = \sqrt{5.88} \approx 2.424 \text{ m/s} \]

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Key Concepts

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

kinematics
Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. In this problem, we look at the motion of the woman as she leaps vertically. Key components include her initial and final positions.

To break down her motion, we begin by identifying the important heights her center of mass reaches: starting at 0.40 meters, then 0.90 meters when her feet leave the floor, and finally 1.20 meters at the peak of her leap.

By converting these heights from centimeters to meters, we use clear and concise measurements, which are crucial in physics calculations. This conversion helps in applying consistent units when calculating work, force, and energy later on.
conservation of energy
The conservation of energy principle states that energy cannot be created or destroyed, only converted from one form to another. This problem illustrates the conversion between kinetic energy and potential energy during the woman's leap.

Initially, as the woman crouches, her potential energy is low. When she starts to leap, her muscles do work to increase both her kinetic and potential energies. The work done by the woman is equal to the change in her potential energy from 0.40 meters to 1.20 meters.

The formula used here is:




We calculate this work to be 431.2 joules. Energy conservation is also used to determine her maximum speed, where all the kinetic energy when her feet leave the floor (at 0.90 meters) is converted to potential energy at the top of her leap (1.20 meters). Therefore, conservation of energy helps us solve for her velocity using the formula derived from kinetic and potential energy equality.
forces and Newton's laws
Forces and Newton's laws are fundamental in analyzing any motion. Newton's second law states that the force acting on an object equals its mass times its acceleration (F = ma).

In this problem, we need to find the average force the ground exerts on the woman. The work-energy principle, a derivative of Newton's laws, is applied here. The work done on the woman (431.2 joules) translates to the force applied over the distance her center of mass moves while she presses down (0.50 meters).

Using the formula for work (W = Fd), we rearrange it to find the average force:

This gives us an average force of 862.4 newtons. Newton's laws not only help in understanding the forces involved but also in relating them to motion through quantities like work and energy.
gravitational potential energy
Gravitational potential energy (GPE) is the energy an object possesses because of its height above the ground. The formula for GPE is
.

In this exercise, the woman鈥檚 gravitational potential energy changes as she moves upward. Initially, when crouched at 0.40 meters, her potential energy is relatively low. As she leaps to 1.20 meters, it increases.

The change in potential energy is calculated as:

.

Using the woman's mass (55 kg) and the height change (from 0.40m to 1.20m), we calculate the work done as 431.2 joules. This example of gravitational potential energy highlights the direct link between height and energy in physics problems.

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Most popular questions from this chapter

Neutron Scattering Show that if a neutron is scattered through \(90^{\circ}\) in an elastic collision with an initially stationary deuteron, the neutron loses \(\frac{2}{3}\) of its initial kinetic energy to the deuteron. (In atomic mass units, the mass of a neutron is \(1.0 \mathrm{u}\) and the mass of a deuteron is \(2.0 \mathrm{u}\).)

Elastic Collision of Cart A cart with mass \(340 \mathrm{~g}\) moving on a frictionless linear air track at an initial speed of \(1.2 \mathrm{~m} / \mathrm{s}\) undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at \(0.66 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the two-cart center of mass?

Steel Ball and Block A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 10-61). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

Spring at the Top of an Incline a spring with spring constant \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a \(37.0^{\circ}\) frictionless incline. The lower end of the incline is \(1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

Potential Energy Function A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=(-4.00 \mathrm{~J} / \mathrm{m}) e^{(-x /(4.00 \mathrm{~m}))} $$ At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) and (d) the greatest value of \(x\) between which the particle can move. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine the equation for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0 ?\)
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