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Chapter 5: Problem 52

An \(85 \mathrm{~kg}\) man lowers himself to the ground from a height of \(10.0 \mathrm{~m}\) by holding onto a rope that runs over a frictionless pulley to a \(65 \mathrm{~kg}\) sandbag. With what speed does the man hit the ground if he started from rest?

Short Answer

Expert verified
The man hits the ground at a speed of approximately 5.12 m/s.

Step by step solution

01

Understand the Problem

This problem involves a man and a sandbag connected over a pulley. The man lowers himself from a certain height, and we need to determine the speed at which he reaches the ground. Both the man and the sandbag are subject to gravitational forces.
02

Identify the Forces Involved

The two main forces are the gravitational forces acting on the man and the sandbag. The man has a gravitational force of \( F_{m} = 85 ext{ kg} \times 9.8 ext{ m/s}^2 \) and the sandbag has \( F_{s} = 65 ext{ kg} \times 9.8 ext{ m/s}^2 \).
03

Set Up the Equations of Motion

Use Newton's second law to set up the equations. The net force on the system contributes to the acceleration. The difference in the gravitational forces of the man and the sandbag determine the net force: \( F_{ ext{net}} = F_{m} - F_{s} \). This net force equals \( (m + s) imes a \) where \( m \) and \( s \) are masses of the man and the sandbag respectively.
04

Calculate the Net Force and Acceleration

Compute the net force: \( F_{ ext{net}} = (85 ext{ kg} \times 9.8 ext{ m/s}^2) - (65 ext{ kg} \times 9.8 ext{ m/s}^2) = 196 ext{ N} \). The mass of the system is \( 85 ext{ kg} + 65 ext{ kg} = 150 ext{ kg} \). The acceleration \( a \) then is \( \frac{196 ext{ N}}{150 ext{ kg}} = 1.31 ext{ m/s}^2 \).
05

Apply Kinematic Equation

Use the kinematic equation \( v^2 = u^2 + 2as \) to find the final speed, where initial speed \( u = 0 \), \( a = 1.31 ext{ m/s}^2 \), and displacement \( s = 10.0 ext{ m} \).
06

Calculate the Final Speed

Substitute the known values into the equation: \( v^2 = 0 + 2 \times 1.31 ext{ m/s}^2 \times 10.0 ext{ m} \). Thus, \( v^2 = 26.2 \), leading to \( v = \sqrt{26.2} = 5.12 ext{ m/s} \).

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Key Concepts

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

Gravitational Force
Gravitational force is the force by which a planet or other celestial body attracts objects towards its center. It's what keeps us grounded on Earth. The formula to calculate gravitational force is given by:
  • \( F = m \cdot g \)
Here,
  • \( F \) is the gravitational force.
  • \( m \) is the mass of the object and is measured in kilograms (\( ext{kg} \)).
  • \( g \) is the acceleration due to gravity, which on Earth is approximately \( 9.8 \, ext{m/s}^2 \).

In this exercise, both the man and the sandbag experience gravitational forces.
For the man: \( F_m = 85 \, ext{kg} \times 9.8 \, ext{m/s}^2 = 833 \, ext{N} \)
For the sandbag: \( F_s = 65 \, ext{kg} \times 9.8 \, ext{m/s}^2 = 637 \, ext{N} \)
These forces are crucial in determining the net force in our pulley system exercise.
Newton's Second Law
Newton's second law of motion tells us how the motion of an object changes when it is acted upon by external forces. It is often stated as:
  • \( F_{\text{net}} = m \cdot a \)
Where:
  • \( F_{\text{net}} \) is the net force applied to the system.
  • \( m \) is the total mass of the system.
  • \( a \) is the acceleration of the system.

In our pulley system, to find the acceleration, we first need the net force.
The two gravitational forces are acting in opposite directions, giving us:
  • \( F_{\text{net}} = F_m - F_s \)
This results in a net force pushing the man downward.
Then, we calculate the acceleration of the system:
  • \( a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{196 \, ext{N}}{150 \, ext{kg}} = 1.31 \, ext{m/s}^2 \)
Newton's second law helps us determine how these forces translate into movement.
Kinematic Equations
Kinematic equations are used to describe the motion of objects. They allow us to calculate changes in speed, position, and time for an object in motion when the acceleration is constant. The key kinematic equation used in this exercise is:
  • \( v^2 = u^2 + 2as \)
Where:
  • \( v \) is the final velocity we're solving for.
  • \( u \) is the initial velocity, which is \( 0 \) since the man starts from rest.
  • \( a \) is the acceleration we calculated, \( 1.31 \, ext{m/s}^2 \).
  • \( s \) is the displacement, which is \( 10.0 \, ext{m} \).
Substituting the known values into the equation gives:
  • \( v^2 = 0 + 2 \times 1.31 \, ext{m/s}^2 \times 10.0 \, ext{m} = 26.2 \)
Solving for \( v \), we find the man's speed when he hits the ground is approximately \( 5.12 \, ext{m/s} \). This equation lets us see how his acceleration and initial rest state lead to his final speed.

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Most popular questions from this chapter

While two forces act on it, a particle is to move at the constant velocity \(\vec{v}=(3 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(4 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). One of the forces is \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((-6 \mathrm{~N}) \hat{\mathrm{j}} .\) What is the other force?

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A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

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