Integral problem question (energy and momentum)

In summary: So 0= 2a+ b so b= -2a and F(x)= ax- 2a. But you are given that! F(x) is linear because "F decreases uniformly from 2 N to 0 over 2 meters". Like any linear function, F can be written as F(x)= ax+ b and you know thatF(0)= 10, F(2)= 0. So 0= 2a+ b so b= -2a and F(x)= ax- 2a.In summary, the conversation is about a problem involving integrations and differentiations. The person is having trouble with part b of the question, which asks to find the
  • #1
Yura
39
0
i got a sheet full of a lot of questions on integrations and differentiations and got them all except for two of them. in both the questions i run into the same type of problem so if i get one then i can probably work out the other using the same method.
here's the question that's troubling me:

(i've worked out part a) but i can't seem to get out part b). )

a particle with mass 80g is acted on by a force which decreases uniformly with respect to displacement from 10N to zero over 2 metres.

a) calculate the maximum velocity of the particle, given v(0) = 0.

b) find the time for which the force becomes zero.


the part a) answer that i got was 9m/s.
for the part b) its like I am missing some information but seeing as there's two questions like that on the sheet I am guessing there's a method to do this. what i think i need is a function of the Force to x so that i can use
dx/dt = dx/dF * dF/dt.
but unless i have that infomation, right now I am clueless on how to continue.


EDIT: I've done a little more on it but i ended up with the integral of
(-62.5x^2 + 250x)^(-1-2) dx ... but i don't know how to do this integration.

much appreciated if someone can show me how this is done.
thanks.
 
Last edited:
Physics news on Phys.org
  • #2
" what i think i need is a function of the Force to x so that i can use
dx/dt = dx/dF * dF/dt."

But you are given that! F(x) is linear because "F decreases uniformly from 2 N to 0 over 2 meters".
Like any linear function, F can be written as F(x)= ax+ b and you know that
F(0)= 10, F(2)= 0.
 
  • #3


It seems like you are trying to solve an integral problem involving energy and momentum. It's great that you were able to solve part a) of the problem and find the maximum velocity of the particle. However, in order to solve part b), you need to understand a few concepts related to energy and momentum.

Firstly, you need to understand that the force acting on the particle is related to its acceleration through the equation F=ma, where m is the mass of the particle and a is its acceleration. In this problem, the force is decreasing uniformly from 10N to 0N over a distance of 2 meters. This means that the acceleration of the particle is also changing uniformly over this distance.

Now, to find the time for which the force becomes zero, you need to use the concept of work and energy. The work done by a force is equal to the change in kinetic energy of the particle. In this case, the initial kinetic energy of the particle is zero and the final kinetic energy is given by 1/2 mv^2 (where v is the velocity of the particle). Therefore, using the work-energy theorem, you can set up the following equation:

Work done by the force = Change in kinetic energy
=> Force x Displacement = 1/2 mv^2

Substituting the given values, we get:

10N x 2m = 1/2 (0.08kg) v^2
=> v^2 = 40/0.08 = 500
=> v = √500 = 22.36 m/s

So, the maximum velocity of the particle is 22.36 m/s. Now, to find the time for which the force becomes zero, you can use the equation of motion:

v = u + at

where u is the initial velocity (which is 0 in this case), a is the acceleration (which is changing uniformly from 10N to 0N over a distance of 2 meters), and t is the time for which the force becomes zero.

Therefore, we can write:

22.36 = 0 + (10/0.08)t
=> t = 22.36 x 0.08/10 = 0.1789 seconds

So, the time for which the force becomes zero is approximately 0.1789 seconds.

In order to solve the integration problem, you need to use the concept of work and energy
 

FAQ: Integral problem question (energy and momentum)

What is the "integral problem" in energy and momentum?

The "integral problem" in energy and momentum refers to the challenge of solving for the total energy and momentum of a system when only certain components or variables are known. It involves using mathematical techniques, such as integration, to find the missing information.

How is the integral problem related to energy and momentum conservation?

The integral problem is closely related to energy and momentum conservation, as it involves determining the total energy and momentum of a system, which must be conserved according to the laws of physics. Solving the integral problem allows us to understand how energy and momentum are transferred and conserved within a system.

What are some common approaches to solving the integral problem in energy and momentum?

There are several common approaches to solving the integral problem in energy and momentum. These include the use of mathematical equations, such as the integral form of the conservation of energy and momentum equations, as well as graphical methods, such as energy and momentum diagrams.

Can the integral problem be applied to real-world scenarios?

Yes, the integral problem can be applied to real-world scenarios in a variety of fields, including physics, engineering, and even economics. It is a fundamental concept in understanding how energy and momentum behave in different systems and can be used to solve practical problems and make predictions.

What are some potential limitations or challenges of solving the integral problem in energy and momentum?

One potential limitation of solving the integral problem in energy and momentum is the complexity of the systems being studied. In some cases, the equations and methods used to solve the problem may not accurately represent the real-world scenario, leading to potential errors. Additionally, the integral problem may become more challenging to solve when considering multiple interacting systems or when dealing with non-conservative forces.

Similar threads

Replies
6
Views
2K
Replies
1
Views
1K
Replies
10
Views
2K
Replies
5
Views
2K
Replies
4
Views
1K
Back
Top