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, 4.1 m) while a constant force acts on it. The force has magnitude 2.4 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?
___________J
work= force*distnace * cosine anglebetween
To calculate the work done by a force during a displacement, we need to use the formula:
Work = Force * Displacement * cos(theta),
where:
- Force is the magnitude of the force,
- Displacement is the magnitude of the displacement,
- theta is the angle between the force vector and displacement vector.
In this case, the force magnitude is given as 2.4 N, and the displacement is from the origin to a point at (3.0 m, 4.1 m). We can calculate the displacement magnitude as follows:
Displacement Magnitude = sqrt((3.0 m)^2 + (4.1 m)^2)
= sqrt(9.0 m^2 + 16.81 m^2)
= sqrt(25.81 m^2)
= 5.08 m.
The angle between the force and displacement vectors is given as 100° counterclockwise from the positive x-axis. However, we need to convert it to radians for the trigonometric function. To convert degrees to radians, we use the formula:
theta (rad) = theta (degrees) * (pi / 180).
theta (rad) = 100° * (pi / 180)
= 1.75 rad.
Now, we can substitute the values into the work formula:
Work = 2.4 N * 5.08 m * cos(1.75 rad)
= 12.192 N*m * cos(1.75 rad)
= 12.192 N*m * 0.9998477
≈ 12.19 J.
Therefore, the work done by the force on the coin during the displacement is approximately 12.19 J.