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(II) How much tension must a cable withstand if it is used to accelerate a \(1200-\mathrm{kg}\) car vertically upward at 0.70 \(\mathrm{m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
The cable must withstand a tension of 12612 N.

Step by step solution

01

Identify the Forces Involved

First, determine which forces are acting on the car. There is the gravitational force acting downward, which is the car's weight, and there is the tension in the cable acting upward, which must overcome gravity and provide the net upward acceleration.
02

Calculate the Gravitational Force

The gravitational force can be found using the formula \( F_g = mg \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \mathrm{m/s^{2}} \)). So for a \( 1200 \, \mathrm{kg} \) car: \[ F_g = 1200 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^{2}} = 11772 \, \mathrm{N}. \]
03

Calculate the Net Force Needed for Acceleration

Use Newton's second law \( F = ma \) to calculate the net force required for the car to accelerate at \( 0.70 \, \mathrm{m/s^{2}} \). \[ F_{net} = 1200 \, \mathrm{kg} \times 0.70 \, \mathrm{m/s^{2}} = 840 \, \mathrm{N}. \]
04

Determine the Total Tension in the Cable

The tension in the cable must counteract both the gravitational force and provide the additional force for the acceleration. Thus, the total tension \( T \) is the sum of the gravitational force and the net force for acceleration:\[ T = F_g + F_{net} \]\[ T = 11772 \, \mathrm{N} + 840 \, \mathrm{N} = 12612 \, \mathrm{N}. \]

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Key Concepts

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

Newton's Second Law of Motion
Understanding Newton's second law is crucial to solving problems involving forces and motion. This fundamental law expresses the relationship between the force acting on an object, its mass, and the acceleration it experiences. The law can be stated as:
  • Force = Mass × Acceleration
In mathematical terms: \[ F = ma \] This equation tells us that the force applied to an object is directly proportional to its mass and the acceleration produced. The force is measured in newtons (N), mass in kilograms (kg), and acceleration in meters per second squared (m/s²).In the context of the exercise at hand, we apply this principle to find the net force needed to accelerate the car upward. Once calculated, this net force is added to the gravitational force exerted by the car's weight to ensure the total tension the cable must withstand.
Gravitational Force
Gravitational force acts on all objects with mass, pulling them towards the center of the Earth. It’s most commonly referred to as "weight" in everyday usage. This force can be calculated using the formula: \[ F_g = mg \] where
  • \( F_g \) is the gravitational force,
  • \( m \) is the mass of the object (in kg),
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \mathrm{m/s^{2}} \) on the surface of the Earth.
In our exercise, the car's mass is known to be \( 1200 \, \mathrm{kg} \). Plugging these values into the formula gives us the gravitational force exerted by the car. This is a key step in finding the total tension in the cable, as this force needs to be counteracted by the tension for the car to move upward.
Acceleration
Acceleration describes the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. The unit of acceleration is meters per second squared (\( \mathrm{m/s^{2}} \)).In cases where an object is accelerating upwards, such as in this exercise, it's essential to calculate the net force required to achieve this acceleration. This involves using Newton's second law, as shown earlier, to determine how much additional force is necessary to overcome not just inertia, but also any opposing forces like gravity.In the given problem, the car needs to be accelerated at \( 0.70 \, \mathrm{m/s^{2}} \). Given the car's mass, we find the required net force needed to achieve this acceleration using the formula:\[ F_{net} = ma \]This gives us a part of the total tension calculation as it reveals how much extra force the cable must exert in addition to balancing the gravitational force.

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Most popular questions from this chapter

A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g's. Calculate the force on a \(65-\mathrm{kg}\) person accelerating at this rate. What distance is traveled if brought to rest at this rate from \(95 \mathrm{~km} / \mathrm{h} ?\)

(II) At the instant a race began, a \(65-\mathrm{kg}\) sprinter exerted a force of 720 \(\mathrm{N}\) on the starting block at a \(22^{\circ}\) angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? \((b)\) If the force was exerted for \(0.32 \mathrm{s},\) with what speed did the sprinter leave the starting block?

A particular race car can cover a quarter-mile track \((402 \mathrm{~m})\) in \(6.40 \mathrm{~s}\) starting from a standstill. Assuming the acceleration is constant, how many "g's" does the driver experience? If the combined mass of the driver and race car is \(535 \mathrm{~kg},\) what horizontal force must the road exert on the tires?

What is the weight of a \(68-\mathrm{kg}\) astronaut \((a)\) on Earth, (b) on the Moon \(\left(g=1.7 \mathrm{~m} / \mathrm{s}^{2}\right),(c)\) on Mars \(\left(g=3.7 \mathrm{~m} / \mathrm{s}^{2}\right),\) (d) in outer space traveling with constant velocity?

A \(75.0-\mathrm{kg}\) person stands on a scale in an elevator. What does the scale read (in \(\mathrm{N}\) and in \(\mathrm{kg}\) ) when \((a)\) the elevator is at rest, (b) the elevator is climbing at a constant speed of \(3.0 \mathrm{~m} / \mathrm{s},(c)\) the elevator is descending at \(3.0 \mathrm{~m} / \mathrm{s},(d)\) the elevator is accelerating upward at \(3.0 \mathrm{~m} / \mathrm{s}^{2},(e)\) the elevator is accelerating downward at \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

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