# Work Kinetic energy using variable force

• kjamha
In summary, the conversation discusses the position of an object with a changing mass and its velocity and work done by forces acting on it. The solution for part (a) is correct, and for part (b), the velocity at time 0 is -6 m/s. The acceleration is found by taking the derivative, and if the force is 0, then the velocity at time 0 would also be 0.
kjamha

## Homework Statement

The position of an object of mass 2 kg is changing as a function of time by the formula x(t) = 4t^3-6t. (a) find the work done by the forces acting on the particle between t=1 and t=3s. (b) What is the object's velocity when the force on it is zero?

## Homework Equations

is my work for part (a) correct?
I do not know how to approach b.

## The Attempt at a Solution

I took the derivative of the equation
12(t)^2 - 6 at t=1s and t=3s
to find vi and vf (I have vi = 6 m/s and vf = 102 m/s).
Then I used W = change in KE
1/2 x 2(102^2) - 1/2 x 2 (6^2) = 10,368 J.
Is that correct?
I'm not sure how to start part b

And the time derivative of velocity is?

Are you asking for the acceleration?

kjamha said:
Are you asking for the acceleration?
Yes. Acceleration. Does that help for part b?

kjamha said:

## Homework Equations

is my work for part (a) correct?
I do not know how to approach b.

kjamha, your Relevant equations aren't equations. You should place equations that you think are relevant to the type of problem being solved in that portion of the template.

taking the derivative, I would get acceleration = 24t
if F = 0, then 0 = 24t and t would have to be 0
find the velocity at time 0
12(t)^2 - 6 = -6

gneill said:
kjamha, your Relevant equations aren't equations. You should place equations that you think are relevant to the type of problem being solved in that portion of the template.
ok - I read it quickly and thought it said relevant questions.

kjamha said:
taking the derivative, I would get acceleration = 24t
if F = 0, then 0 = 24t and t would have to be 0
find the velocity at time 0
12(t)^2 - 6 = -6
Yes, that's correct.

PeroK said:
Yes, that's correct.
Thank you!

## 1. What is work kinetic energy using variable force?

Work kinetic energy using variable force is a measure of the change in an object's kinetic energy due to a variable force acting on it. It takes into account both the magnitude and direction of the force, as well as the distance over which the force is applied.

## 2. How is work kinetic energy using variable force calculated?

The formula for work kinetic energy using variable force is W = ∫F(x)dx, where W represents work, F(x) represents the variable force as a function of distance, and dx represents an infinitesimal change in distance. This formula can be found by taking the integral of the force over the distance traveled.

## 3. What is the relationship between work and kinetic energy using variable force?

The work done by a variable force is equal to the change in kinetic energy of an object. This means that when work is done on an object by a variable force, the object's kinetic energy will either increase or decrease depending on the direction of the force.

## 4. How does work kinetic energy using variable force differ from work kinetic energy using constant force?

The main difference between these two concepts is that work kinetic energy using variable force takes into account the changing force over a distance, while work kinetic energy using constant force assumes a constant force acting on the object. This means that the calculations for work kinetic energy will be different in each case.

## 5. What are some real-world examples of work kinetic energy using variable force?

Examples of work kinetic energy using variable force include throwing a ball, pushing a shopping cart, and pedaling a bike. In each of these cases, the force acting on the object changes as it moves, resulting in a change in the object's kinetic energy.

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