# Work done by a Variable Force

• mbrmbrg
In summary, the force from 0 to 3.0 seconds causes a change in the position of the object of 7.0 meters.

#### mbrmbrg

A single force acts on a 7.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 3.0 s.

I want to use the equation $$W = \int F(x)dx = \int madx$$
To find acceleration, I took the second derivative of the position function:
$$x(t) = 3t-4t^2+t^3$$
$$v=x'(t) = 3-8t+3t^2$$
$$a=x''(t) = -8+6t$$

When I go to plug values into my work integral, I get $$W=\int m(6t-8)dx$$
I would then take mass out of the integral, but having t and dx in the integral is evil. Do I say that dx=x'(t) and so substitute my velocity function (with dt tacked onto the end) for dx?

Last edited:
"Yes". The change of variable theorem states that,

$$\int_{x(t=0)}^{x(t=3)}F(x)dx = \int_0^3F(x(t))\dot{x}(t)dt$$

But as you cleverly noted, F(x(t))=ma(t).

mbrmbrg said:
A single force acts on a 7.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 3.0 s.

I want to use the equation $$W = \int F(x)dx = \int madx$$
To find acceleration, I took the second derivative of the position function:
$$x(t) = 3t-4t^2+t^3$$
$$v=x'(t) = 3-8t+3t^2$$
$$a=x''(t) = -8+6t$$

When I go to plug values into my work integral, I get $$W=\int m(6t-8)dx$$
I would then take mass out of the integral, but having t and dx in the integral is evil. Do I say that dx=x'(t) and so substitute my velocity function (with dt tacked onto the end) for dx?
That looks like it will take away the evil and give you the correct solution!l

Thank you very kindly!

## 1. What is work done by a variable force?

Work done by a variable force is the product of the force applied on an object and the displacement of the object in the direction of the force. Unlike a constant force, a variable force changes in magnitude and/or direction during the motion of the object.

## 2. How is the work done by a variable force calculated?

The work done by a variable force is calculated by integrating the force function with respect to displacement. This means finding the area under the force-displacement curve. Mathematically, it can be expressed as W = ∫F(x)dx, where W is work, F(x) is the force function, and dx is the infinitesimal displacement.

## 3. What is the difference between positive and negative work done by a variable force?

Positive work is done when the force and displacement are in the same direction, meaning the force is causing the object to move. Negative work is done when the force and displacement are in opposite directions, meaning the force is resisting the motion of the object. The total work done by a variable force is the sum of positive and negative work.

## 4. Can the work done by a variable force be negative?

Yes, the work done by a variable force can be negative. As mentioned before, negative work occurs when the force and displacement are in opposite directions. This can happen when the force is slowing down or stopping the motion of an object, or when the object is moving in the opposite direction of the force.

## 5. What are some real-life examples of work done by a variable force?

Examples of work done by a variable force include pushing or pulling a shopping cart, riding a bike, or lifting weights. In each of these cases, the force applied on the object varies depending on the position and movement of the object. Other examples can include opening a door, swinging on a swing, or throwing a ball.