Why do you integrate expressions for work + (hw problems)?

In summary, the work done by the conservative force acting on a 5.00-kg particle moving from x=1.00m to x=5.00m, described by the equation Fx = (2x + 4) N, is 40J. This can be calculated by taking the integral of the force with respect to distance, using the correct limits for the integral. In general, integration is necessary when the force applied varies with distance.
  • #1
Struggle Help
2
0
PHYSICS FORUMS RULES REQUIRE ONLY ONE PROBLEM POSTED PER THREAD.
Question:
A single conservative force acts on a 5.00-kg particle. The equation Fx = (2x + 4) N, where x is in
meters, describes this force. As the particle moves along the x-axis from x = 1.00 m to x = 5.00 m,
calculate (a) the work done by this force.

Work = Force * Distance.

(2*(5-1) +4) = 12N
12N * 4 = 48J.. (The correct answer is 40J).


If we integrate the equation (?) it results in (4)^2 + 4(4) = 32J (Still wrong answer).

First of all how do you get the right answer?

Second of all (more conceptual) why do you (in general) have to integrate work expressions? In what other situations do you have to integrate equations/ expressions?


Question 2:
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591
N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is
the same during starting and stopping, and determine (a) the weight of the person, (b) the person’s mass,
and (c) the acceleration of the elevator.

a) 491 N
b)50.1kg
c)2m/s^2

a) We can assume the person's weight didn't change, and the magnitude of acceleration is the same during stopping and starting (albeit in different directions I'm assuming).

When the elevator starts 3 forces are affecting the scale, Fg(gravity), Fn (normal force), Fe(force of elevator). When the elevator starts Fe is 100N downwards, and when the elevators stops Fe is 100N upwards. How would I in this situation determine the weight in Newtons of the person? Take the average of the two forces, during stopping and starting?

b) 491N / 9.8m/s^2 = 50/1kg

c)Not sure how to start this.

Question 3:
The Earth rotates about its axis with a period of 24.0 h. Imagine that the rotational speed can be
increased. If an object at the equator is to have zero apparent weight, (a) what must the new period be?
(b) By what factor would the speed of the object be increased when the planet is rotating at the higher
speed?

a)1.41h

b)17.1

This question is really challenging (for me at least) so I really just expect for the teacher to go over it in class. I would appreciate it if you could patch up my misunderstandings in the first two more basic problems I have posted.

This is my first time on this forum so please forgive me for any mistakes or transgressions I may have committed. I would really appreciate any help you may offer me.
 
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  • #2
Struggle Help said:
Question:
A single conservative force acts on a 5.00-kg particle. The equation Fx = (2x + 4) N, where x is in
meters, describes this force. As the particle moves along the x-axis from x = 1.00 m to x = 5.00 m,
calculate (a) the work done by this force.

Work = Force * Distance.

(2*(5-1) +4) = 12N
12N * 4 = 48J.. (The correct answer is 40J).

If we integrate the equation (?) it results in (4)^2 + 4(4) = 32J (Still wrong answer).

First of all how do you get the right answer?

You aren't using the correct limits for your integral, which is why your calculation of work is wrong. Work = integral of Force w.r.t. distance.

Second of all (more conceptual) why do you (in general) have to integrate work expressions?

In this particular case, the amount of force varies with the distance x. If the force were constant, an integral would not be necessary.

In what other situations do you have to integrate equations/ expressions?

It depends. There is no way to answer this vague question.
 
  • #3
SteamKing said:
You aren't using the correct limits for your integral, which is why your calculation of work is wrong. Work = integral of Force w.r.t. distance.

That makes sense, thanks.

I redid the integral, trying x*(X+4)

5*(5+4) - 1*(1+4) = 40. Thanks!
 

Related to Why do you integrate expressions for work + (hw problems)?

1. Why do you integrate expressions for work?

Integrating expressions for work allows us to calculate the total amount of work done by a variable force over a certain distance. This is particularly useful in physics and engineering applications where the force may change continuously.

2. What is the purpose of integrating expressions for work?

The purpose of integrating expressions for work is to determine the total amount of energy expended in performing a certain task. This can help us understand the efficiency of a system or the amount of energy required to complete a specific task.

3. How does integrating expressions for work differ from other methods of calculation?

Integrating expressions for work takes into account the variable force applied over a certain distance, while other methods of calculation may only consider a constant force or a specific point in time. This makes integration a more accurate and comprehensive method of determining work.

4. Can you provide an example of integrating expressions for work in practice?

Sure! For example, if you are pushing a box with a variable force of 10N over a distance of 5 meters, you can use integration to calculate the total work done. This would involve finding the area under the curve of the force-distance graph, which would give you the total work done.

5. How is integrating expressions for work related to solving HW problems?

Integrating expressions for work is a fundamental concept in physics, and many homework problems involve calculating work done by varying forces. Understanding how to integrate these expressions allows you to solve these problems accurately and efficiently.

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