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, 5.8 m) while a constant force acts on it. The force has magnitude 2.8 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?
work is force in direction of motion * distance moved
so we need that angle between the motion and the force
For the force:
Tan theta = 5.8/3
theta = 62.65 degrees from x axis
now the force is 100 deg from x axis so
angle between force and motion = 100 - 62.65 = 37.35
so
component of F in direction of motion =
2.8 cos 37.35 = 2.23 Newtons
D = distance moved = sqrt (9+5.8^2)
= 6.53 meters
so finally
work done = F dot D = 2.23*6.53
=14.6 Joules
To find the work done by a force on an object, you can use the formula:
Work = Force * Displacement * cos(theta)
where:
- Work is the work done by the force
- Force is the magnitude of the force
- Displacement is the displacement vector
- theta is the angle between the force vector and the displacement vector
In this case, the force is 2.8 N and it is directed at an angle of 100° counterclockwise from the positive direction of the x-axis. The displacement vector is the vector from the origin (0, 0) to the point (3.0 m, 5.8 m).
Now, let's find the components of the displacement vector. The x-component is 3.0 m and the y-component is 5.8 m.
Using the formula for work, we can calculate:
Work = 2.8 N * [(3.0 m * cos(100°)) + (5.8 m * cos(90°))]
Let's break it down:
- The x-component of the displacement vector (3.0 m) is multiplied by the cosine of the angle (100°) because the force acts in the x direction.
- The y-component of the displacement vector (5.8 m) is multiplied by the cosine of 90° because the force does not act in the y direction.
Now, let's calculate:
Work = 2.8 N * [(3.0 m * cos(100°)) + (5.8 m * cos(90°))]
Work = 2.8 N * [(3.0 m * -0.173648) + (5.8 m * 0)]
Work = 2.8 N * [(3.0 m * -0.173648)]
Work = 2.8 N * (-0.520944 m)
Work = -1.461 m * N
Therefore, the work done by the force on the coin during the displacement is -1.461 m * N.