Q. A force F = (3.99xi + 2.86yj) N acts on an object as it moves in the x direction from the origin to x = 4.45 m. Calculate the work done on the object by the force.

Attemp:
since we only consider x direction i multiplied 3.99 by 4.45 but it didn't work out.

work done is integral Fdotdx

INT (3.99x)dx= 1/2 3.99x^2 over limits, or

work= 1/2 3.99 (4.45)^2

I am assuming you are taking calculus.

Thank You!!!

To calculate the work done on an object by a force, you need to use the equation:

Work = Force * Distance * cos(theta)

Where:
- Work is the work done on the object (in joules, J)
- Force is the applied force (in newtons, N)
- Distance is the distance over which the force is applied (in meters, m)
- Theta is the angle between the force and the direction of motion (in radians, rad)

In this case, the force vector given is F = 3.99xi + 2.86yj N. Since the object is moving only in the x-direction, the angle "theta" between the force and the direction of motion is 0 degrees or 0 radians.

Now, we can calculate the work done on the object as it moves in the x-direction from the origin to x = 4.45 m:

Work = (Force_x * Distance * cos(theta))

Since we only consider the x-component of the force, we can calculate the work as:

Work = (3.99 N * 4.45 m * cos(0))

Cancelling out the cosine of 0, we get:

Work = 3.99 N * 4.45 m

Therefore, the work done on the object by the force is:

Work = 17.7455 J

So the work done on the object by the force is approximately 17.7455 joules (J).