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kondor19780726 [428]

4 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?

2 answers:

Serggg [28]4 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]4 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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I hope it helps you!

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