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?

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

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HugoYates HugoYates · Answer 2 · 2019-03-30T11:37:31+01:00

Answer

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,[tex]F_{x}[/tex] = [tex]2.5Ncos100[/tex]

=[tex]2.5\times (-.173)[/tex]

[tex]F_{x}[/tex]=-0.43

component of force in y direction,[tex]F_{y}[/tex] = [tex]2.5sin 100[/tex]

=[tex]2.5\times .98[/tex]

[tex]F_{y}[/tex]=2.46N

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

W=[tex]F_{x}\times x+F_{y}\times y[/tex]

W = [tex]-0.43\times 3m + 2.46N\times 4[/tex]

=[tex]-1.29+9.8[/tex]

W=8.55Joule