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Chapter 3: Problem 49

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 with a speed of 5.11 m/s.

Step by step solution

01

Analyze the system

Consider the man and the sandbag as an interacting system. Set the man鈥檚 descent and the sandbag鈥檚 ascent as positive directions.
02

Determine the force equations

Apply Newton鈥檚 second law to the man and the sandbag. For the man: \[ m_{\text{man}} g - T = m_{\text{man}} a \] For the sandbag: \[ T - m_{\text{sandbag}} g = m_{\text{sandbag}} a \]
03

Solve for tension and acceleration

Solve the system of equations: Add the two equations: \[ m_{\text{man}} g - m_{\text{sandbag}} g = (m_{\text{man}} + m_{\text{sandbag}}) a \] Isolate the acceleration: \[ a = \frac{(m_{\text{man}} - m_{\text{sandbag}})g}{(m_{\text{man}} + m_{\text{sandbag}})} \] Plug in the known values: \[ a = \frac{(85 \text{ kg} - 65 \text{ kg}) \times 9.81 \text{ m/s}^2}{(85 \text{ kg} + 65 \text{ kg})} = \frac{20 \times 9.81}{150} = 1.308 \text{ m/s}^2 \]
04

Use kinematic equations

To find the final velocity, use the kinematic equation: \[ v_f = v_i + a t \] and \[ d = v_i t + \frac{1}{2} a t^2 \] Since the initial velocity \((v_i)\) is zero: \[ d = \frac{1}{2} a t^2 \] Rearrange to solve for time \(t\): \[ t^2 = \frac{2d}{a} \] \[ t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2 \times 10}{1.308}} = \sqrt{15.29} = 3.91 \text{ s} \]
05

Calculate final velocity

Use the time found to calculate the final velocity: \[ v_f = a t = 1.308 \text{ m/s}^2 \times 3.91 \text{ s} = 5.11 \text{ m/s} \]

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Key Concepts

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

Force Analysis
Understanding force analysis is crucial in solving physics problems involving motion. Force analysis helps us identify and break down the forces acting on objects. In this exercise, we have two objects: an 85 kg man and a 65 kg sandbag. The forces include:
  • Gravity acting on both the man: \( m_{\text{man}} g \) and the sandbag: \( m_{\text{sandbag}} g \)
  • The tension in the rope (T) opposing gravity for the man and pulling up the sandbag
By setting the man's descent and the sandbag's ascent as positive, we align their motions and apply Newton's Second Law for both objects. This gives us the equations:
\[ m_{\text{man}} g - T = m_{\text{man}} a \] \[ T - m_{\text{sandbag}} g = m_{\text{sandbag}} a \]
These equations set up the system for solving the problem.
Kinematic Equations
Kinematic equations describe the motion of objects under constant acceleration. They link variables such as displacement (d), initial velocity (v_i), final velocity (v_f), acceleration (a), and time (t). Common kinematic equations include:
\[ v_f = v_i + a t \] \[ d = v_i t + \frac{1}{2} a t^2 \]
In this problem, the initial velocity (v_i) is zero because the man starts from rest. Thus, the second equation simplifies to:
\[ d = \frac{1}{2} a t^2 \] By rearranging it to solve for t, we find how long it takes for the man to reach the ground.
Acceleration Calculation
Calculating acceleration is necessary to understand how the velocity of the objects changes over time. From the force analysis equations, adding both equations gives:
\[ m_{\text{man}} g - m_{\text{sandbag}} g = (m_{\text{man}} + m_{\text{sandbag}}) a \] Solving for acceleration (a) involves isolating a:
\[ a = \frac{(m_{\text{man}} - m_{\text{sandbag}}) g}{(m_{\text{man}} + m_{\text{sandbag}})} \]
Plugging in the values:
\[ a = \frac{(85 \text{ kg} - 65 \text{ kg}) \times 9.81 \text{ m/s}^2}{(85 \text{ kg} + 65 \text{ kg})} = 1.308 \text{ m/s}^2 \]
This tells us how quickly the speeds of the man and the sandbag change.
Tension in Ropes
The tension in the rope (T) is the force transmitted through the rope that affects both objects. It opposes gravity for the man and assists in lifting the sandbag. By solving the equations:
For the man: \[ T = m_{\text{man}} g - m_{\text{man}} a \] For the sandbag: \[ T = m_{\text{sandbag}} g + m_{\text{sandbag}} a \] These expressions help us understand how tension balances and changes due to the mass and acceleration of the objects.
Final Velocity Computation
Calculating final velocity tells us how fast the man is moving when he reaches the ground. Once we know the time (t), we use the kinematic equation:
\[ v_f = v_i + a t \] Given the initial velocity (v_i) is zero:
\[ v_f = a t \] With a = 1.308 m/s^2 and t = 3.91 s:
\[ v_f = 1.308 \text{ m/s}^2 \times 3.91 \text{ s} = 5.11 \text{ m/s} \] This final velocity indicates how fast the man hits the ground, helping to understand the result of forces and motion involved.

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Most popular questions from this chapter

Raindrops fall \(1700 \mathrm{~m}\) from a cloud to the ground. (a) If they were not slowed by air resistance, how fast would the drops be moving when they struck the ground? (b) Would it be safe to walk outside during a rainstorm?

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

A physics student is standing on one of the steps of the fire ladder behind a building on campus doing a physics experiment. From there she drops a stone (without giving it any initial velocity) and notes that it takes approximately \(2.45 \mathrm{~s}\) to hit the ground. The second time she throws the stone vertically upward and notes that it takes approximately \(5.16 \mathrm{~s}\) for it to hit the ground. (a) Calculate the height above the parking lot from which she releases the first stone. (b) Calculate the initial velocity with which she has thrown the second stone upward. (c) How high above the parking lot did the second stone rise before it started falling again?

In order to minimize the \(g\) forces on you, suppose you decide to accelerate with a constant acceleration such that you reach half the speed of light \((c / 2=1.5 \times\) \(10^{8} \mathrm{~m} / \mathrm{s}\) ) at the midpoint of your trip and then start slowing down so you are at rest just in time to dock with the supply vessel at the other end of this solar system. (a) Draw a single motion diagram showing the speeding-up and slowing-down processes. (b) In a coordinate system in which you move along the positive \(x\) axis, what is the direction and magnitude of your initial acceleration? In other words, is your acceleration positive or negative? (c) In a coordinate system in which you move along the positive \(x\) axis, what is the direction and magnitude of your acceleration while you are slowing down for your rendezvous with the supply vessel? In other words, is your acceleration positive or negative? (Hint: The answer to part (b) and a symmetry argument can save you some effort.) (d) How long will your overall trip take? (e) If the Defiant has a mass of \(M=2.850 \times 10^{8} \mathrm{~kg}\), what is the thrust force (in newtons) needed to accelerate your starship? (f) The amount of force you feel being impressed on you by the back of your seat as the starship picks up speed is proportional to your acceleration. A common way to measure typical forces you might feel is to calculate \(g\) -forces. This is done by comparing the acceleration you experience to the acceleration you would experience while falling freely close to the surface of the Earth. Thus, you can find \(g\) -forces by dividing your acceleration by \(9.8 \mathrm{~m} / \mathrm{s}^{2}\). What \(g\) -forces would you experience while accelerating in the Defiant? (g) The maximum sustained \(g\) -force that a human can stand is about \(3 \mathrm{~g}\). What would happen to you during your leisurely acceleration to half the speed of light?

Your roommate peeks over your shoulder while you are reading a physics text and notices the following sentence: "In free fall the acceleration is always \(g\) and always straight downward regardless of the motion." Your roommate finds this peculiar and raises three objections: (a) If I drop a balloon or a feather, it doesn't fall nearly as fast as a brick. (b) Not everything falls straight down; if I throw a ball it can go sideways. (c) If I hold a wooden ball in one hand and a steel ball in the other, I can tell that the steel ball is being pulled down much more strongly than the wooden one. It will probably fall faster. How would you respond to these statements? Discuss the extent to which they invalidate the quoted statement. If they don't invalidate the statement, explain why.
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