Solving Force Problems: Work Done from x=0 to x=4m

[brad sue]

Hi, I try to solve the following problem but my answer does not correspond to the solution of the textbook. Please help me to understand the problem:
The net force acting on a particle depends on the position of the particle along the x-axis according the relation F=F₀+C*x, where F₀=5N and C=-2N/m. The particle is initially at rest at x=0m when the force begins to act.
a- Calculate the work done by the force when the particle reaches x values 1,2,3and 4 m?
b- Determineany positions ( besides x=0m ) where the work done is zero.

I have problem for the question a.
What I did is to compute F₁=F₀+C*(1) [for x=1m]
that gives me F₁=5-2=3N
So the work for F₁ is W₁= F₁*(1)=3J
However the answer in the textbook it 4J I don't know how they get that.
Thank you for your time.
B

 

[Diane_]

You're neglecting the fact that the force is not constant over any distance.

If you know calculus, you can do this by integrating the force over the distance.

If you don't, then consider: since the force changes linearly, you could calculate an average force over a given distance without loss of rigor.

One thing, whichever way you do it - note the negative on your constant. What does that mean physically?

 

[quicknote]

Hi there,

Firstly, I want to commend you for seeking help and trying to understand the problem better. As a scientist, it is important to always question and seek clarification when things don't make sense.

Now, let's break down the problem and the solution to help you understand it better.

The problem gives us the relationship between force (F) and position (x) as F=F0+C*x, where F0 is the initial force and C is a constant. This means that as the particle moves along the x-axis, the force acting on it changes according to this relationship.

For part a, we are asked to calculate the work done by the force when the particle reaches different positions (1, 2, 3, and 4m). To calculate work, we use the formula W=F*d, where W is work, F is the force, and d is the displacement. So, for each position, we need to calculate the force at that position and multiply it by the displacement (in this case, the displacement is 1m).

So, for x=1m, the force would be F1=F0+C*x = 5N + (-2N/m)*(1m) = 3N. This is the correct answer, as you have calculated. The work done at this position would be W1=F1*d=3N*1m=3J.

For x=2m, the force would be F2=F0+C*x = 5N + (-2N/m)*(2m) = 1N. So, the work done at this position would be W2=F2*d=1N*1m=1J.

Similarly, for x=3m, the force would be F3=F0+C*x = 5N + (-2N/m)*(3m) = -1N. And the work done at this position would be W3=F3*d=-1N*1m=-1J.

Finally, for x=4m, the force would be F4=F0+C*x = 5N + (-2N/m)*(4m) = -3N. And the work done at this position would be W4=F4*d=-3N*1m=-3J.

So, the work done at x=1m is indeed 3J, and the work done at x=4m is -3J (which is equivalent to