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An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of \(4.5 \times 10^{4} \mathrm{kg},\) and the force causes its speed to change from 7100 to \(5500 \mathrm{m} / \mathrm{s}\). (a) What is the work done by the force? (b) If the asteroid slows down over a distance of \(1.8 \times 10^{6} \mathrm{m},\) determine the magnitude of the force.

Short Answer

Expert verified
(a) The work done is \(-3.2 \times 10^{10}\) J. (b) The force magnitude is \(-1.78 \times 10^{4}\) N.

Step by step solution

01

Understanding the Problem

We have an asteroid with a mass of \(4.5 \times 10^4\) kg. Its speed changes from 7100 m/s to 5500 m/s. We need to find the work done by the force that slows it down and the magnitude of the force, given that the slowing distance is \(1.8 \times 10^6\) m.
02

Calculate Initial and Final Kinetic Energy

The initial kinetic energy (KE) of the asteroid is \(KE_i = \frac{1}{2} m v_i^2\). Substituting the values, \(KE_i = \frac{1}{2} \times 4.5 \times 10^4 \times 7100^2\). For the final kinetic energy, \(KE_f = \frac{1}{2} m v_f^2\), substituting the values, \(KE_f = \frac{1}{2} \times 4.5 \times 10^4 \times 5500^2\).
03

Calculate the Work Done by the Force

Work done by the force is equal to the change in kinetic energy.\[W = KE_f - KE_i\]. Calculate \(KE_i\) and \(KE_f\) using the values from Step 1 and find \(W\).
04

Use the Work-Energy Principle to Determine Force

The work done is also defined as \(W = F \cdot d\), where \(F\) is the force and \(d = 1.8 \times 10^6\) m. From the calculated work done in Step 2, solve for the force using \(F = \frac{W}{d}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a way to measure an object's energy based on its motion. The formula used to calculate the kinetic energy \( KE \) of an object is \( KE = \frac{1}{2} mv^2 \), where \( m \) represents the mass in kilograms and \( v \) is the velocity in meters per second.
In the case of the asteroid, we start by calculating its initial and final kinetic energy. Initially, the velocity \( v_i \) is 7100 m/s, which means we need to substitute this into the formula along with the mass \( 4.5 \times 10^4 \) kg. This calculation provides us with the initial kinetic energy:
\[ KE_i = \frac{1}{2} \times 4.5 \times 10^4 \times 7100^2. \]
For the final kinetic energy, we use the final velocity \( v_f \) of 5500 m/s in a similar manner:
\[ KE_f = \frac{1}{2} \times 4.5 \times 10^4 \times 5500^2. \]
The difference between these two energy values tells us about the work done by the force in changing the asteroid's energy state.
Force Calculation
Force calculations often begin with the concept of work done, which in physics is a way to quantify the energy transferred when an object is moved over a distance by a force. Here, the work done \( W \) is found using the change in kinetic energy. Using the work-energy principle, we know
\[ W = KE_f - KE_i. \]
This principle states that the total work done on an object is equal to the change in its kinetic energy.
Once we've determined the work \( W \), we can find the force \( F \) applied using the formula
\[ W = F \cdot d, \]
where \( d \) is the distance over which the force is applied, which is given as \( 1.8 \times 10^6 \) m in this asteroid scenario. Rearranging the formula as
\[ F = \frac{W}{d}, \]
allows us to solve for the magnitude of the force. This gives us a clear understanding of how much force is required to slow the asteroid over the specified distance.
Asteroid Motion
Asteroids are fascinating celestial objects orbiting around the sun, mostly found in the asteroid belt between Mars and Jupiter. Understanding their motion is crucial for navigational and collision avoidance studies.
In this exercise, the asteroid's motion through space is influenced by a force that alters its speed. Initially, the asteroid was moving at 7100 m/s, and due to the applied force, slows down to 5500 m/s. This change in motion implies the asteroid is experiencing a deceleration.
Such deceleration (slowing down) might be due to gravitational pulls from nearby planets, or possibly due to impacts with other space debris. The change in speed over a particular distance (1.8 million meters in this case) gives us key insights into the forces acting upon it.
Knowing how to calculate these changes in motion is crucial in the broader study of asteroids, particularly when planning missions to potentially redirect their paths or understanding their origins and compositions.

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Most popular questions from this chapter

A 55.0-kg skateboarder starts out with a speed of 1.80 \(\mathrm{m} / \mathrm{s}\). He does \(+80.0 \mathrm{J}\) of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{m} / \mathrm{s}\). (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?

In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide down an icy track, belly down and head first. In the 2010 Winter Olympics, the track had sixteen turns and dropped \(126 \mathrm{m}\) in elevation from top to bottom. (a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed at the beginning of the run is relatively small and can be ignored. (b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat with a speed of \(40.5 \mathrm{m} / \mathrm{s}\) (about \(91 \mathrm{mi} / \mathrm{h}\) ). How much work was done on him and his sled (assuming a total mass of \(118 \mathrm{kg}\) ) by nonconservative forces during this heat?

Multiple-Concept Example 5 reviews many of the concepts that play roles in this problem. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of \(25.0^{\circ}\) with the horizontal. The coefficient of kinetic friction between her skis and the snow is \(0.200 .\) She coasts down a distance of \(10.4 \mathrm{m}\) before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is \(3.50 \mathrm{m}\) below the edge. How fast is she going just before she lands?

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta\). One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0} .\) On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

A \(1.00 \times 10^{2}-\mathrm{kg}\) crate is being pushed across a horizontal floor by a force \(\overrightarrow{\mathbf{P}}\) that makes an angle of \(30.0^{\circ}\) below the horizontal. The coefficient of kinetic friction is \(0.200 .\) What should be the magnitude of \(\overrightarrow{\mathbf{P}},\) so that the net work done by it and the kinetic frictional force is zero?

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