A \(20 \mathrm{~kg}\) chimp climbs three meters at a constant speed along a rope in 4 seconds. What is the power the chimp expends against gravity? (A) \(100 \mathrm{~W}\) (B) \(150 \mathrm{~W}\) (C) \(200 \mathrm{~W}\) (D) \(300 \mathrm{~W}\) (E) \(600 \mathrm{~W}\)

Short Answer

Expert verified
A) 100 W B) 150 W C) 200 W D) 250 W Answer: B) 150 W

Step by step solution

01

Calculate the force exerted by the chimp - its weight

To find the force exerted by the chimp against gravity, we need to find its weight. Weight can be calculated using the equation Weight = mass × acceleration due to gravity, where the mass of the chimp is 20 kg and acceleration due to gravity (g) is approximately 9.8 meters per second squared (m/s^2). Weight = (20 kg)(9.8 m/s^2) = 196 N
02

Calculate the work done against gravity

To find the work done against gravity, we need to multiply the force exerted by the chimp with the distance it climbed. In this case, the distance climbed is 3 meters. Work = Force × Distance = (196 N)(3 m) = 588 J (Joules)
03

Calculate the power expended against gravity

Now we can find the power expended by dividing the work done by the time taken to climb. The chimp took 4 seconds to climb the distance. Power = Work / Time = (588 J) / (4 s) = 147 W (Watts)
04

Choose the correct answer

Now we can compare the calculated power with the available options. The closest value to our result is 150 W. The answer is (B) 150 W.

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Key Concepts

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

Power Expenditure Calculation
Understanding how to calculate power expenditure is crucial in physics, especially when discussing work and energy. To delve into this concept, let's consider the example of a chimp climbing a rope. The power exerted is a measure of how quickly work is done, and in this scenario, it involves the chimp working against the force of gravity.

To calculate power, we first need to establish the work done. In the case of the chimp, the work done is the product of the force, which is the weight of the chimp due to gravity, and the distance climbed. Since power is the rate at which work is done, we divide the total work by the time taken to climb.

It's essential to remember the units when calculating power: work is in Joules and time in seconds, so power is in Watts. Power expenditure is a key concept connecting work and time, making it invaluable in various areas such as mechanics and energetics.
Work-Energy Principle
The work-energy principle is a foundational concept in physics that relates the work done on an object to its change in energy. When discussing this principle, it's essential to note that work done against a force, such as gravity, results in a change in the object's potential energy.

Here's a closer look at the problem of the chimp climbing a rope. The chimp does work against gravity, which translates to an increase in gravitational potential energy as the chimp ascends. By calculating this work, we obtain the amount of energy transferred in the process.

In this context, work is not lost but stored as potential energy, a concept highlighting the conservation of energy. The work-energy principle is a powerful tool in problem-solving because it provides a direct link between force and energy, allowing us to analyze various physical situations with precision.
Force and Motion
Force and motion are inseparable aspects of physics. Newton's second law of motion states that the force exerted on an object is equal to the mass of the object times its acceleration (\( F = ma \)). In our chimp problem, the force exerted is due to gravity, acting to pull the chimp downwards, and is calculated as the product of the mass and the acceleration due to gravity.

However, because the chimp moves at a constant speed, the actual motion does not involve acceleration. Instead, this scenario involves overcoming a constant force over a distance, resulting in work being done. The chimp's action of climbing reflects a balance of forces where the force exerted by the chimp through the rope equals the gravitational pull.

Understanding force in the context of motion not only aids in problem-solving but is also central to understanding how objects interact within various systems. From celestial mechanics to the smallest atomic scales, force and motion are key to deciphering the dynamics of our universe.

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Most popular questions from this chapter

Consider the same setup as in problem 20 , except that now friction exists between the \(15 \mathrm{~kg}\) block and the table, with a coefficient of kinetic friction \(\mu\). Consider the new tension \(T\) in the string and the new acceleration of the blocks, \(a\). One can conclude: (A) \(T\) and \(a\) are both the same as they were in the case without friction. (B) \(T\) and \(a\) are both larger than they were in the case without friction. (C) \(T\) and \(a\) are both smaller than they were in the case without friction. (D) \(T\) is larger than it was in the case without friction, but \(a\) is smaller. (E) \(T\) is smaller than it was in the case without friction, but \(a\) is larger.

A \(1 \mathrm{~kg}\) mass strikes a \(2 \mathrm{~kg}\) mass in an inelastic collision on a horizontal frictionless surface. The \(1 \mathrm{~kg}\) mass is traveling at \(1 \mathrm{~m} / \mathrm{s}\) in the positive direction and the \(2 \mathrm{~kg}\) mass is traveling at \(1 \mathrm{~m} / \mathrm{s}\) in the negative direction. The final velocity of the masses is (A) \(+1 \mathrm{~m} / \mathrm{s}\) (B) \(+1 / 2 \mathrm{~m} / \mathrm{s}\) (C) \(-1 \mathrm{~m} / \mathrm{s}\) (D) \(-2 / 3 \mathrm{~m} / \mathrm{s}\) (E) \(-1 / 3 \mathrm{~m} / \mathrm{s}\)

Two planets, Gamma and Delta, with masses \(m_G\) and \(m_D\) and radii \(R_G\) and \(R_D\) are initially at rest a large distance \(d\) apart from each other. Khan the Conqueror suddenly allows the two planets to approach under their mutual gravitational attraction until they collide. To determine the velocity at which Gamma and Delta collide you could (A) use conservation of energy alone. (B) use conservation of momentum alone. (C) use both conservation of energy and momentum. (D) The problem cannot be solved using energy and momentum conservation. (E) The problem cannot be solved unless you know how long it takes the two planets to collide.

Two objects, \(m\) and \(M\), with initial velocities \(v_0\) and \(U_0\) respectively, undergo a one-dimensional elastic collision. Show that their relative speed is unchanged during the collision. (The relative velocity between \(m\) and \(M\) is defined as \(v-U_{-}\))

An object moving at a constant velocity is acted on by several forces. Which of the following is true? i. All the forces are acting in the same or opposite directions. ii. The vector sum of the forces must be zero. iii. The magnitude of the forces must be the same. (A) i (B) ii (C) iii (D) i and iii (E) i, ii and iii

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