A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (3.0 m, 4.4 m) while a constant force acts on it. The force has magnitude 2.3 N and is directed at a counterclockwise angle of 100° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

Force vector, F:

(2.3cos(100°),2.3sin(100°)) N
Displacement vector, D:
(3,4.4) m
Work done = F.D (dot product of the two vectors) J

A dot product between two vectors A(ax,ay) and B(bx,by) is a scaler calculated as
A.B = ax*bx + ay*by
See also:
http://en.wikipedia.org/wiki/Dot_product

To calculate the work done by a force, you need to use the formula:

Work = Force * Displacement * cos(theta)

where:
- Force is the magnitude of the force (2.3 N in this case),
- Displacement is the distance the coin moves from the origin to the point (3.0 m in the x-direction and 4.4 m in the y-direction),
- theta is the angle between the force vector and the displacement vector (in this case, 100° from the positive x-axis counterclockwise).

To find the displacement, we can use the Pythagorean theorem since the displacement is given in the x and y directions:

Displacement = sqrt(dx^2 + dy^2)

where:
- dx is the displacement in the x-direction (3.0 m),
- dy is the displacement in the y-direction (4.4 m).

Let's calculate the displacement:

Displacement = sqrt((3.0 m)^2 + (4.4 m)^2)
= sqrt(9.0 m^2 + 19.36 m^2)
= sqrt(28.36 m^2)
= 5.33 m (rounded to two decimal places)

Now, we can calculate the work done by the force:

Work = (2.3 N) * (5.33 m) * cos(100°)

To calculate the cosine of 100°, we need to convert it to radians:

100° * (π/180°) ≈ 1.75 rad

Work = (2.3 N) * (5.33 m) * cos(1.75 rad)
≈ 25.12 J (rounded to two decimal places)

Therefore, the work done by the force on the coin during the displacement is approximately 25.12 Joules.