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Chapter 4: Problem 75

The parachute on a race car of weight \(8.820 \mathrm{~N}\) opens at the end of a quarter-mile run when the car is traveling at \(35 \mathrm{~m} / \mathrm{s} .\) What total retarding force must be supplied by the parachute to stop the car in a distance of \(1000 \mathrm{m?}\)

Short Answer

Expert verified
The total retarding force supplied by the parachute must be \(0.54 \mathrm{~N}\)

Step by step solution

01

Calculate the mass of the car

We can calculate the mass of the car by rearranging the formula for weight (Weight = Mass * Gravity) to solve for mass (Mass = Weight / Gravity). Given the weight of the car as \(8.820 \mathrm{~N}\) and the gravity as \(9.8 \mathrm{~m/s^2}\), the mass of the car would be \(m = 8.820 \mathrm{~N} / 9.8 \mathrm{~m/s^2} = 0.9 \mathrm{~kg}\).
02

Calculate the deceleration of the car

We can calculate the deceleration of the car using the final velocity, initial velocity and the distance over which the deceleration occurs. The formula we will use is \( v^2 = u^2 + 2as \) where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration (deceleration in this case) and \( s \) is distance. Rearranging this formula for \( a \), we get \( a = (v^2 - u^2) / (2s) \). Plugging in the given values, \( v = 0 \mathrm{~m/s} \) (as the car comes to a stop), \( u = 35 \mathrm{~m/s} \) and \( s = 1000 \mathrm{~m} \), we calculate \( a = -(35 \mathrm{~m/s})^2 / (2 * 1000 \mathrm{~m}) = -0.6 \mathrm{~m/s^2} \). Here, the negative sign indicates that this is a deceleration.
03

Calculate total retarding force

We can use Newton's second law of motion formula (Force = Mass * Acceleration) to calculate the retarding force required to stop the car. Plugging in the mass of the car (\(0.9 \mathrm{~kg}\)) and the deceleration we calculated (-0.6 \(\mathrm{m/s^2}\)), the retarding force that must be supplied by the parachute is \( F = 0.9 \mathrm{~kg} * -0.6 \mathrm{~m/s^2} = -0.54 \mathrm{~N} \), where the negative sign indicates that it is a retarding force

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Key Concepts

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

Deceleration
Deceleration refers to the process of slowing down an object. When the velocity of an object decreases, we say the object is experiencing deceleration. In physics, deceleration can be calculated using the equation:
  • Initial velocity (\( u \))
  • Final velocity (\( v \))
  • Acceleration (negative in the case of deceleration, \( a \))
  • Distance (\( s \)) over which the change occurs
The relevant formula is:
\[ v^2 = u^2 + 2as \]
However, when dealing with deceleration, \( a \) becomes negative. To find deceleration, we rearrange it to:
\[ a = \frac{v^2 - u^2}{2s} \]
In this specific case, where the car starts at 35 m/s and needs to stop at a final velocity (\( v \)) of 0 m/s over a distance of 1000 m, the deceleration would be:
\[-0.6 \, m/s^2\].
Here, the negative sign clearly illustrates the slowing down process.
Newton's Second Law
Newton's Second Law of Motion offers a fundamental explanation about how forces affect motion. This law states that the force (\( F \)) needed to accelerate or decelerate an object is directly proportional to the mass (\( m \)) of the object and the acceleration (\( a \)) it undergoes. Mathematically, it can be represented as:
\[ F = m \cdot a \]
When applying this to our situation, the mass of the car (\( 0.9 \, kg \)) and the calculated deceleration (\(-0.6 \, m/s^2\)) provide the retarding force. According to the laws of motion, the result is:
\[-0.54 \, N\].
The negative sign is crucial as it confirms the force is applied in the opposite direction of motion, emphasizing its role as a retarding force to stop the car smoothly.
Retarding Force
Retarding force is another term for a force that reduces the velocity of an object. It's effectively a force that opposes the motion of an object. It's the counterpart of the familiar concept of acceleration, but works to decelerate instead.
In our car's situation, the retarding force is generated by the parachute, intending to halt the car's forward motion. This force acts opposite to the car's moving direction.
The magnitude of the retarding force can be determined by applying Newton鈥檚 Second Law, where the mass (\( m \)) of the object and the deceleration (\( a \)) are key components:
\[ F = m \cdot a \]
The result of \(-0.54 \, N\) shows us that the parachute's deploying force is just enough to stop the car within the given distance. It's important to remember that negative signs often denote direction in physical contexts, so while 'negative' might sound bad, it simply means the force is opposite to the motion in this case.

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Most popular questions from this chapter

Objects of masses \(m_{1}=4.00 \mathrm{~kg}\) and \(m_{2}=9.00 \mathrm{~kg}\) are connected by a light string that passes over a frictionless pulley as in Figure \(\mathrm{P} 4.54\). The object \(m_{1}\) is held at rest on the floor, and \(m_{2}\) rests on a fixed incline of \(\theta=40.0^{\circ}\). The objects are released from rest, and \(m_{2}\) slides \(1.00 \mathrm{~m}\) down the incline in \(4.00 \mathrm{~s}\). Determine (a) the acceleration of each object, (b) the tension in the string, and (c) the coefficient of kinetic friction between \(m_{2}\) and the incline.

A coin is placed near one edge of a book lying on a table, and that edge of the book is lifted until the coin just slips down the incline as shown in Figure \(\mathrm{P} 4,70\). The angle of the incline, \(\theta_{c}\), called the critical angle, is measured. (a) Draw a frec-body diagram for the coin when it is on the verge of slipping and identify all forces acting on it. Your free- body diagram should include a force of static friction acting up the incline. (b) Is the magnitude of the friction force equal to \(\mu_{3} n\) for angles less than \(\theta_{c}\) ? Explain. What can you definitely say about the magnitude of the friction force for any angle \(\theta \leq \theta_{c} ?\) (c) Show that the coefficient of static friction is given by \(\mu_{\mathrm{s}}=\tan \theta_{c}\). (d) Once the coin starts to slide down the incline, the angle can be adjusted to a new value \(\theta_{c}^{\prime} \leq \theta_{c}\) such that the coins moves down the incline with constant speed. How does observation enable you to obtain the coefficient of kinetic friction?

A \(1000-\mathrm{kg}\) car is pulling a \(300-\mathrm{kg}\) trailer. Together, the car and trailer have an acceleration of \(2.15 \mathrm{~m} / \mathrm{s}^{2}\) in the forward direction. Neglecting frictional forces on the trailer, determine (a) the net force on the car, (b) the net force on the trailer, (c) the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road.

As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). In the process, he moves his hand through a distance of \(1.50 \mathrm{~m}\). If the hall has a mass of \(0.150 \mathrm{~kg}\), find the force he exerts on the ball to give it this upward speed.

(a) An clevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?^{2}\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w ?\) (c) When the elevator is moving upward, but the acceleration is downward, which is greater, Tor w? (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\), Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find \(T\). Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\), Find \(T\). Is your answer consistent with your answer to part (c)?
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