Question

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.50 m, 5.30 m) while a constant force acts on it. The force has magnitude 7.40 N and is directed at a counterclockwise angle of 104.° from the positive direction of the x axis. How much work is done by the force on the coin during the displacement?

Answer 1

Answer:

The work done by the force is 35.37 Joules.

Explanation:

Given that,

coin started sliding from origin (0,0) to point (1.50m,5.30m)

therefore displacement of coin along x-axis,x = 1.50m

displacement of coin along y-axis,y = 5.30m

angle between applied force and positive x-axis = 104

applied force = 7.40N

Now we will find

component of force along x-axis

$F_{x} = 7.4Ncos104F_{x}=-1.7902$

component of force in y direction,

$F_{y} = 7.4sin 104F_{y}=7.1082N$

work is done by the force on the coin during the displacement will be given as

$W=F_{x}\times x+F_{y}\times yW = -1.7902\times 1.5m + 7.1802N\times 5.3W=35.37Joule$

So, the work done by the force is 35.37 Joules.