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kondor19780726 [428]
3 years ago
6

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?
Physics
2 answers:
Serggg [28]3 years ago
5 0
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
Lera25 [3.4K]3 years ago
3 0

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,F_{x} = 2.5Ncos100

=2.5\times (-.173)

F_{x}=-0.43

component of force in y direction,F_{y} = 2.5sin 100

=2.5\times .98

F_{y}=2.46N  

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

W=F_{x}\times x+F_{y}\times y

W = -0.43\times 3m + 2.46N\times 4

=-1.29+9.8

W=8.55Joule

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