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CP You are pushing a large box across a frictionless floor by applying a constant horizontal force. If the box starts at rest, you have to do work \(W_{1}\) in order for the box to travel a distance \(d\) in time \(t .\) How much work would you have to do, in terms of \(W_{1}\), to make the box go the same distance in half the time?

Short Answer

Expert verified
In order to make the box go the same distance in half the time, you would have to do 4 times the initial amount of work, i.e., \(W_2 =4 W_1\).

Step by step solution

01

Deriving the original velocity

As the box starts from rest and moves with constant force, it also implies constant acceleration. Using the kinematic equations of motion, we can get the velocity after time t in terms of distance travelled: \(v = \sqrt{2ad}\)
02

Calculating the kinetic energy

The work done on the box is equal to the change in kinetic energy (K.E.). Since the box starts from rest, the initial kinetic energy is zero. Therefore, the work done, \(W_1\) is equal to the final kinetic energy, which can be calculated using the formula for kinetic energy: \(K.E. = 1/2 m v^2\). From Step 1, we substitute \(v = \sqrt{2ad}\) to find \(W_1 = m.a.d\) since \(K.E. = m.a.d\)
03

Calculating the velocity for the same distance in half the time

Now to cover the same distance d in half the time (t/2), we can use the kinematic equations of motion to get the new velocity \(v' = 2\sqrt{2ad}\)
04

Calculating the new amount of work

The new amount of work \(W_2\) will again be equal to the final kinetic energy which is now \(K.E.' = 1/2 m v'^2\). Substituting \(v' = 2\sqrt{2ad}\) to find the new amount of work: \(W_2 = m . a . d = 4W_1\).

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Key Concepts

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

Work-Energy Principle
The Work-Energy Principle is a core concept in physics that describes the relationship between work and energy in mechanical systems. It states that the work done on an object is equal to the change in its kinetic energy. In simple terms, whenever work is done on an object, there will be a corresponding change in the object's kinetic energy.

Considering our exercise where a box is pushed across a floor, the principle is applied to determine how much work is needed to move the box a certain distance. When you apply a constant horizontal force to the box, the work done on the box, denoted as \(W_{1}\), results in an increase in the box's kinetic energy from zero, as it starts at rest. This change in kinetic energy can be calculated using the formula \(K.E. = \frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) is the velocity of the box. By understanding this principle, we can predict and calculate the work needed under different conditions, such as moving the box faster over the same distance.
Kinematic Equations
Kinematic Equations are used to describe the motion of objects without taking into account the forces that cause the motion. These equations provide a link between an object's displacement, velocity, acceleration, and the time of travel. For objects with constant acceleration, we often use the equation \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.

In the provided exercise, kinematic equations allow us to derive the velocity of the box after a certain time, given its constant acceleration. The solution uses these equations to find the initial and the new velocity when the box is made to travel the same distance in half the time. By mastering kinematic equations, students can predict how the velocity of an object changes with time—which is critical to solving problems involving motion.
Kinetic Energy
Kinetic Energy (K.E.) is the energy that an object possesses due to its motion. It's given by the equation \(K.E. = \frac{1}{2} m v^2\), where \(m\) is the mass of the object, and \(v\) is its velocity. An important aspect of kinetic energy is that it is proportional to the square of the velocity, meaning that even a small increase in velocity leads to a significant increase in kinetic energy.

Referring to our exercise, the kinetic energy equation is used to determine the work done on the box, \(W_1\), to reach a certain velocity after traveling distance \(d\) under constant acceleration. When the box is required to travel the same distance in half the time, it must reach a greater velocity, which in turn means its kinetic energy increases—and therefore, the work done, denoted as \(W_2\), must also increase. This conceptual understanding of kinetic energy enables us to foresee how changes in velocity affect the energy and work associated with an object's motion.
Constant Acceleration
Constant Acceleration occurs when an object's velocity changes at a steady rate over time. This is often due to a constant unbalanced force acting on the object as per Newton's second law of motion. When dealing with constant acceleration, the motion can be fully described by the kinematic equations, which simplify the analysis of such systems.

In the context of our exercise, the box starts from rest and is pushed with a constant horizontal force, leading to constant acceleration. This allows the use of kinematic equations to calculate the box’s final velocity after a certain time, and when the time is halved to maintain the same distance traveled. Understanding constant acceleration is vital as it helps to predict how the speed of an object changes over time, which is essential when determining the work required to modify the object's state of motion.

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Most popular questions from this chapter

A \(75.0 \mathrm{~kg}\) painter climbs a ladder that is \(2.75 \mathrm{~m}\) long and leans against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

You push your physics book \(1.50 \mathrm{~m}\) along a horizontal tabletop with a horizontal push of \(2.40 \mathrm{~N}\) while the opposing force of friction is \(0.600 \mathrm{~N}\). How much work does each of the following forces do on the book: (a) your \(2.40 \mathrm{~N}\) push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

A spring of force constant \(300.0 \mathrm{~N} / \mathrm{m}\) and unstretched length \(0.240 \mathrm{~m}\) is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to \(15.0 \mathrm{~N}\). How long will the spring now be, and how much work was required to stretch it that distance?

You and your bicycle have combined mass \(80.0 \mathrm{~kg}\). When you reach the base of a bridge, you are traveling along the road at \(5.00 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.74\) ). At the top of the bridge, you have climbed a vertical distance of \(5.20 \mathrm{~m}\) and slowed to \(1.50 \mathrm{~m} / \mathrm{s}\). Ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

When its \(75 \mathrm{~kW}\) (100 hp) engine is generating full power, a small single-engine airplane with mass \(700 \mathrm{~kg}\) gains altitude at a rate of \(2.5 \mathrm{~m} / \mathrm{s}(150 \mathrm{~m} / \mathrm{min},\) or \(500 \mathrm{ft} / \mathrm{min}) .\) What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

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