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
Take the "dot product" of the force vector and the displacement vector. If i is the x unit vector and j is the y unit vector, the force vector is
F = 2.4 sin 100 i + 2.4 cos 100 j
The displacement vector is
deltaR = 3.0 i + 4.1 j
W = F . (deltaR)
= 7.2 sin 100 + 9.84 cos 100
To determine the amount of work done by the force on the coin during the displacement, we can use the formula for work:
Work = Force * Displacement * cos(theta)
Where:
- Force is the magnitude of the force acting on the coin (2.4 N),
- Displacement is the distance and direction the coin travels (from the origin to the point with coordinates (3.0 m, 4.1 m)),
- Theta is the angle between the force vector and the displacement vector.
First, let's calculate the displacement:
Displacement = √[(Δx)² + (Δy)²]
Where:
- Δx is the change in x-coordinate (3.0 m - 0 m = 3.0 m),
- Δy is the change in y-coordinate (4.1 m - 0 m = 4.1 m).
Displacement = √[(3.0 m)² + (4.1 m)²] = √[9.0 m² + 16.81 m²] = √25.81 m² = 5.08 m
Next, let's calculate the angle theta:
Theta = 180° - angle
Since the angle is 100° counterclockwise from the positive x-axis, the angle will be:
Theta = 180° - 100° = 80°
Now, we can calculate the work done by the force:
Work = (2.4 N) * (5.08 m) * cos(80°)
Using a calculator, we find:
Work ≈ (2.4 N) * (5.08 m) * 0.17365
Work ≈ 2.0655 J
Therefore, the work done by the force on the coin during the displacement is approximately 2.0655 J.