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In a performance test, each of two cars takes \(9.0 \mathrm{s}\) to accelerate from rest to \(27 \mathrm{m} / \mathrm{s}\). Car A has a mass of \(1400 \mathrm{kg},\) and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{kg} .\) Find the net average force that acts on each car during the test.

Short Answer

Expert verified
Car A: 4200 N, Car B: 5700 N.

Step by step solution

01

Understand the Information Given

We are given two cars that both accelerate from rest to a final velocity of \(27 \mathrm{m/s}\) in \(9.0 \mathrm{s}\). Their masses are \(1400 \mathrm{kg}\) for Car A and \(1900 \mathrm{kg}\) for Car B. We need to find the net average force acting on each car during this acceleration phase.
02

Calculate the Acceleration

Use the formula for acceleration: \(a = \frac{\Delta v}{\Delta t}\) where \(\Delta v = 27 \mathrm{m/s}\) and \(\Delta t = 9.0 \mathrm{s}\). So, \(a = \frac{27 \mathrm{m/s}}{9.0 \mathrm{s}} = 3.0 \mathrm{m/s^2}\). The acceleration for both cars is \(3.0 \mathrm{m/s^2}\).
03

Use Newton's Second Law to Find Force on Car A

Newton's Second Law states \(F = ma\), where \(m\) is mass and \(a\) is acceleration. For Car A, \(m = 1400 \mathrm{kg}\) and \(a = 3.0 \mathrm{m/s^2}\). Thus, \(F = 1400 \mathrm{kg} \times 3.0 \mathrm{m/s^2} = 4200 \mathrm{N}\).
04

Use Newton's Second Law to Find Force on Car B

Similarly, for Car B, \(m = 1900 \mathrm{kg}\) and \(a = 3.0 \mathrm{m/s^2}\). Thus, \(F = 1900 \mathrm{kg} \times 3.0 \mathrm{m/s^2} = 5700 \mathrm{N}\).
05

Conclusion

The net average force acting on Car A is \(4200 \mathrm{N}\) and on Car B is \(5700 \mathrm{N}\).

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Key Concepts

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

Acceleration
Acceleration is the rate at which an object changes its velocity over time. In simpler terms, it tells us how quickly an object speeds up or slows down. In physics, it is usually represented by the symbol \(a\) and is calculated using the formula: \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the time period over which this change happens.
For instance, if a car goes from a standstill to a speed of \(27 \text{ m/s}\) in \(9\) seconds, the acceleration would be \(3.0 \text{ m/s}^2\). This value is a constant in our exercise, which means both cars experience the same rate of increase in speed during the test.
Understanding acceleration can help us predict how a vehicle will behave when a force is applied. It helps engineers design safer vehicles and allows us to estimate how quickly a moving object can reach a certain speed.
Net Force
Net force is the overall force acting on an object when all the individual forces are combined. In the context of Newton's Second Law, net force is critical as it determines how an object accelerates. The formula is \(F_{\text{net}} = ma\), where \(m\) is mass and \(a\) is acceleration.
For cars in the exercise, the net force is what propels them forward during acceleration. It can vary depending on the car's mass and the applied force. According to Newton's Second Law, the more massive an object is, the more net force will be required to achieve the same acceleration. This is why car B, which is heavier, needs a greater force to reach the same speed as car A.
  • The importance of net force is observed in real-world applications like car performance, spacecraft launch, and more.
  • In our specific case, understanding net force helps calculate the traction forces required by friction to move the vehicle.
Mass
Mass is a measure of the amount of matter in an object, typically measured in kilograms. It is a fundamental property that affects how an object accelerates under a given force, according to Newton’s Second Law. Mass is represented by the symbol \(m\).
In our exercise, Car A has a mass of \(1400 \text{ kg}\) while Car B weighs \(1900 \text{ kg}\). This difference in mass directly impacts the net force required to achieve the same acceleration for both vehicles.
  • A higher mass means more inertia; therefore, more force is needed to change the motion of the object.
  • The relationship between mass and force is crucial in designing efficient transportation systems - heavier loads require stronger engines.
The understanding of mass is not only crucial for physics problems but also in real-world logistics and planning.
Velocity
Velocity describes an object's speed in a certain direction. It is a vector quantity, which means it has both magnitude and direction, contrary to speed, which is scalar and only describes how fast an object moves.
In our exercise, each car accelerates from a velocity of \(0\text{ m/s}\) to \(27\text{ m/s}\) over a period of \(9\text{ s}\). This change in velocity helps us calculate the acceleration. The final velocity in this context is important as it represents the speed both cars aim to achieve in the performance test.
Understanding velocity is crucial in various fields:
  • It helps analyze motion, allowing physicists and engineers to predict future positions of moving objects.
  • In navigation, velocity helps map the exact route and time estimated to reach a destination.
In essence, while speed tells us how fast something is moving, velocity gives us a complete picture of motion, incorporating both speed and direction.

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Most popular questions from this chapter

A golf ball strikes a hard, smooth floor at an angle of \(30.0^{\circ}\) and, as the drawing shows, rebounds at the same angle. The mass of the ball is \(0.047 \mathrm{kg},\) and its speed is \(45 \mathrm{m} / \mathrm{s}\) just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

A lumberjack (mass \(=98 \mathrm{kg}\) ) is standing at rest on one end of a floating log (mass \(=230 \mathrm{kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{m} / \mathrm{s}\) relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

Two particles are moving along the \(x\) axis. Particle 1 has a mass \(m_{1}\) and a velocity \(v_{1}=+4.6 \mathrm{m} / \mathrm{s} .\) Particle 2 has a mass \(m_{2}\) and a velocity \(v_{2}=\) \(-6.1 \mathrm{m} / \mathrm{s} .\) The velocity of the center of mass of these two particles is zero. In other words, the center of mass of the particles remains stationary, even though each particle is moving. Find the ratio \(m_{1} / m_{2}\) of the masses of the particles.

A 60.0-kg person, running horizontally with a velocity of \(+3.80 \mathrm{m} / \mathrm{s}\), jumps onto a \(12.0-\mathrm{kg}\) sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast \(30.0 \mathrm{m}\) on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?

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