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Chapter 3: Problem 61

Just as her parachute opens, a \(60-\mathrm{kg}\) parachutist is falling at a speed of \(50 \mathrm{~m} / \mathrm{s}\). After \(0.80 \mathrm{~s}\) has passed, the chute is fully open and her speed has dropped to \(12.0 \mathrm{~m} / \mathrm{s}\). Find the average retarding force exerted upon the chutist during this time if the deceleration is uniform.

Short Answer

Expert verified
The average retarding force is \(-2850 \mathrm{~N}\).

Step by step solution

01

Understand the Given Data

We know the mass of the parachutist is \(60\, \mathrm{kg}\), initial speed \(v_i = 50 \mathrm{~m/s}\), and final speed \(v_f = 12 \mathrm{~m/s}\). The time duration given is \(0.80 \mathrm{~s}\). Our goal is to find the average retarding force.
02

Calculate Acceleration

First, find the acceleration using the formula \(a = \frac{v_f - v_i}{t}\). Substituting the values, \(a = \frac{12 - 50}{0.8}\).
03

Solve for Acceleration

Perform the calculation for acceleration: \(a = \frac{-38}{0.8} = -47.5 \mathrm{~m/s^2}\). The negative sign indicates a deceleration.
04

Use Newton's Second Law

Newton's second law tells us \(F = ma\), where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. Substitute \(m = 60 \mathrm{~kg}\) and \(a = -47.5 \mathrm{~m/s^2}\) into the equation to find \(F\).
05

Calculate the Retarding Force

Compute \(F = 60 \times (-47.5) = -2850 \mathrm{~N}\). The negative sign denotes the force is in the opposite direction of motion.

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Key Concepts

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

Retarding Force
When a parachutist jumps from a plane, they initially accelerate downwards because of gravity. However, once the parachute opens, the effect of air resistance becomes significant.
Air resistance applies a force in the opposite direction to the motion.
This force is known as the retarding force.
  • The retarding force opposes the fall, reducing the speed of the parachutist.
  • It is calculated using Newton's Second Law: \[ F = ma \]where \( F \) is the retarding force, \( m \) is the mass of the object, and \( a \) is the acceleration (or deceleration in this case).
  • In our exercise, it is found to be \(-2850\;\mathrm{N}\).
  • The force is negative, which signifies its direction is opposite to the motion.
Deceleration
Deceleration occurs when an object is slowing down. It is essentially a negative acceleration.
The speed of the parachutist decreases due to the force exerted by the open parachute.
This decrease in speed is what we quantify as deceleration.
  • To find deceleration, use the formula:\[ a = \frac{v_f - v_i}{t} \]where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time taken.
  • In our case, initial speed \( v_i = 50\; m/s \) and final speed \( v_f = 12\; m/s \), with a time of \(0.8\; s\).
  • Substituting these values gives a deceleration of \(-47.5\; m/s^2\).
  • This negative value indicates the parachutist is slowing down.
Uniform Acceleration
Uniform acceleration refers to a constant rate of change in velocity.
In the parachutist problem, deceleration is believed to take place uniformly.
This means the parachutist slows down at a steady rate.
  • Uniform acceleration allows us to use the basic kinematic equations without complication.
  • The given deceleration of \(-47.5\; m/s^2\) was calculated using the assumption of uniformity, meaning it remains constant throughout the \(0.8\; s\).
  • This consistency simplifies calculations and lets us accurately determine forces and velocity changes over time.
Parachutist Problem
The parachutist problem is a classic example in physics to illustrate concepts of forces, motion, and deceleration.
When analyzing such problems, it's essential to consider:
  • The mass of the parachutist, which affects how forces act upon them.
  • The initial and final speeds to understand velocity changes.
  • The time interval over which deceleration occurs.
  • The application of Newton's Second Law to find the retarding forces acting against the parachutist's fall.
By decomposing the scenario into basic physics principles, students can gain a deeper understanding of how forces influence motion and how an opposing force, like a parachute, counters gravity and decelerates a falling object. This problem provides practical insight into how physics operates in real-life situations.

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Most popular questions from this chapter

A \(600-\mathrm{kg}\) car is coasting along a level road at \(30 \mathrm{~m} / \mathrm{s}\). (a) How large a retarding force (assumed constant) is required to stop it in a distance of \(70 \mathrm{~m} ?(b)\) What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction- there's no sliding. (a) First find the car's acceleration from a constant- \(a\) equation. It is known that \(v_{i x}=30 \mathrm{~m} / \mathrm{s}, v_{f x}=0\), and \(x=70 \mathrm{~m}\). Use \(v_{f x}^{2}=v_{i x}^{2}+2 a x\) to find $$ a=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2} $$ Now write $$ F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN} $$ (b) Assume the force found in ( \(a\) ) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is \(F_{\mathrm{f}}=3858 \mathrm{~N}\). The coefficient of friction is given by \(\mu_{s}=F_{\mathrm{f}} / F_{N}\), where \(F_{N}\) is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore, $$ F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N} $$ so that $$ \mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66 $$ The coefficient of friction must be at least \(0.66\) if the car is to stop within \(70 \mathrm{~m}\).

Two cannonballs that each weigh \(4.00 \mathrm{kN}\) on Earth are floating in space far from any other objects. Determine the mutually attractive gravitational force acting on them when they are separated, center-to-center, by \(10.0 \mathrm{~m}\).

Floating in space far from anything else are two spherical asteroids, one having a mass of \(20 \times 10^{10} \mathrm{~kg}\) and the other a mass of \(40 \times 10^{10} \mathrm{~kg}\). Compute the force of attraction on each one due to gravity when their center-to-center separation is \(10 \times 10^{6} \mathrm{~m}\).

A \(400-g\) block with an initial speed of \(80 \mathrm{~cm} / \mathrm{s}\) slides along a horizontal tabletop against a friction force of \(0.70 \mathrm{~N}\). (a) How far will it slide before stopping? \((b)\) What is the coefficient of friction between the block and the tabletop? (a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, \(-0.70 \mathrm{~N}\). Therefore, $$ \sum F=m a \quad \text { becomes } \quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a) $$ from which \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). (Notice that \(m\) is always in kilograms.) To find the distance the block slides, make use of \(v_{i x}=0.80 \mathrm{~m} / \mathrm{s}, v_{f x}=0\), and \(a=-1.75 \mathrm{~m} / \mathrm{s}^{2}\). Then \(v_{f x}^{2}-v_{i x}^{2}=2 a x\) yields $$ x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m} $$ (b) Because the vertical forces on the block must cancel, the upward push of the table \(F_{N}\) must equal the weight \(m g\) of the block. Then $$ \mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18 $$

The fabled planet Dune has a diameter eight times that of Earth and a mass twice as large. If a robot weighs \(1800 \mathrm{~N}\) on the surface of (nonspinning) Dune, what will it weigh at the poles on Earth? Take our planet to be a sphere.
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