Question

A coin Slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates 3.0m,4.0m while a constant force acts on it. The force has magnitude 2.5 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?

Answer 1

x axis :
Fx = 2.5cos(100)
Wx = Fx(X) = 2.5cos(100) x 3
... negative value as Fx has -x direction

y axis :
Fy = 2.5sin(100)
Wy = Fy(Y) = 2.5sin(100) x 5

Workdone = Wx + Wy

Answer 2

W=8.55Joule

Solution

In this question we have given

coin started sliding from origin (0,0) topoint (3m,4m)

therefore displacement of coin along x-axis,x=3m

displacement of coin along y-axis,y=4m

angle between applied force and positive x-axis=100

applied force=2.5N

Now we will find

component of force along x-axis,
`F_(x)` =
`2.5Ncos100`

=
`2.5* (-.173)`

`F_(x)`=-0.43

component of force in y direction,
`F_(y)` =
`2.5sin 100`

=
`2.5* .98`

`F_(y)`=2.46N

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

W=
`F_(x)* x+F_(y)* y`

W =
`-0.43* 3m + 2.46N* 4`

=
`-1.29+9.8`

W=8.55Joule