# Work done by a non-constant force

• clutch12
In summary, the question asks about the separation of two protons that are initially moving towards each other with an initial kinetic energy of 0.18 MeV each. Using the formula for electric potential energy, the final distance between the protons can be calculated by setting the net kinetic energy of the system equal to the potential energy at rest. The final answer is obtained by multiplying the initial kinetic energy by 2 and then solving for the distance.
clutch12

## Homework Statement

Hey this is my first time on here and I'm kinda lost on what I am doing. Any help would be great and the question is
Two protons that are very far apart are hurled straight at each other, each with an initial kinetic energy of 0.18 MeV, where 1 mega electron volt is equal to 1*10^6 multiply (1.6*10^-19) joules. What is the separation of the protons from each other when they momentarily come to a stop?

## Homework Equations

Dont think i can type the equation

## The Attempt at a Solution

So i multiplied the initial kinetic energy by how much 1 electron vold is equal to and then multiplied it by the work

0.18 * 1*10^6 * 1.6* 10^-19
=2.88* 10^-14 J (kinetic energy of both protons)

Next i try to find the distance and this is where i have the trouble

2.88*10^-14 J = (9*10^9)(1.6*10^-19)^2/d
d= (9*10^9)(1.6*10^-19)^2 / 2.88*10^-14 J
d= 8 * 10^-15m

Hi clutch12, welcome to PF.
The system consists of two protons.
Initially what is the total kinetic energy of the system and what is the net potential energy of the system? What are these energies when they are momentarily at rest?

oh so i now need to find the net potential energy of the system right?

and thanks for the welcome

It doesn't make sense to multiply energy by work. The answer would come out in Joules², which doesn't have any meaning for me.

You need a formula for the electric potential energy of a charge near another charge. That is the same as the work done to push one charge close to another. It is the integral of dW = F*dr = kqq/r²*dr
and the answer is kqq/r. No doubt you can look it up in your textbook.
So you have .18 MeV = kqq/r
and you just have to solve for r.

oh never mind you mean kq1q2/r^2 *dr

Last edited:
Delphi51 said:
It doesn't make sense to multiply energy by work. The answer would come out in Joules², which doesn't have any meaning for me.

You need a formula for the electric potential energy of a charge near another charge. That is the same as the work done to push one charge close to another. It is the integral of dW = F*dr = kqq/r²*dr
and the answer is kqq/r. No doubt you can look it up in your textbook.
So you have .18 MeV = kqq/r
and you just have to solve for r.
Both the protons are moving with KE = 0.18 MeV each. So net KE of the system = 2*0.18 MeV.

so basically from what i tried earlier all i would need to do is multiply 0.18 by 2 and then multiply everything all over to get my answer?

and thanks you guys for your help i finally got the answer right

And thanks to rl.bhat for catching the factor of 2!

## What is work done by a non-constant force?

Work done by a non-constant force refers to the amount of energy transferred by a force that changes in magnitude or direction during the process of performing work.

## How is work done by a non-constant force calculated?

The work done by a non-constant force is calculated by finding the area under the force-displacement graph. This can be done by breaking the graph into smaller sections and calculating the area of each section, then adding them together.

## What is the difference between work done by a non-constant force and work done by a constant force?

The main difference is that a constant force maintains the same magnitude and direction throughout the entire process of performing work, while a non-constant force changes in magnitude or direction. This means that the calculation for work done by a constant force is simply force multiplied by displacement, while for a non-constant force, the calculation is more complex.

## How does the angle between the force and displacement affect the work done by a non-constant force?

The angle between the force and displacement affects the work done by a non-constant force because it determines the component of the force that is in the direction of displacement. If the force is perpendicular to the displacement, no work is done. If the force is parallel to the displacement, all of the force contributes to the work done.

## What are some real-life examples of work done by a non-constant force?

Some examples include pushing a shopping cart up a ramp, lifting a bucket of water from a well, and pedaling a bike up a hill. In each of these situations, the force changes in magnitude or direction as the object moves, resulting in a non-constant force and the performance of work.

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