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?

1 answer

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