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).