On a farm, you are pushing on a stubborn pig with a con- stant horizontal force with magnitude 30.0 N and direction 37.0° counterclockwise from the +x-axis. How much work does this force do during a displacement of the pig that is (a) s S = 15.00 m2nd; (b) s S = −16.00 m2ne; (c) s S = −12.00 m2nd + 14.00 m2ne?

The answers to the question are;

(a) 359.4 J

(b) -288.872 J

(c) -34.757 J.

Explanation:

Since the force is acting in the direction 37.0° counterclockwise, we have the horizontal and vertical components, given by

= 30×cos 37° = 23.96 N

= 30×sin 37° = 18.0545 N

Therefore, = (23.96 N) i + (18.0545 N) j

The equation for work done, W by constant force, F on a particle in a straight line is given by

W = F × d × cos θ

Where

d = Distance the object is moved along the straight line by the force, F

θ = Angle between the force and the line of displacement of the object

The equation  F × d × cos θ, is of the form of the scalar product between two vectors that is

Therefore W =   where

= and

=

So that W =  

=  

So we look for the dot product    

(a) When s = 15 m i

=  = and

W =    =  

=  359.4 N·m = 359.4 J

(b) When s = -16 m j

 =   = and

W = =

= -288.872 Nm = -288.872 J

(c) When s = -12.00 m i + 14.00 j, we have

W =   =  -287.52 + 252.763

= -34.757 Nm =

-34.757 J.

Some parts of the question are not well interpreted. The correct interpretation is;

(a) S = (15.00m)i^; (b) S = (−16.00m)j^; (c) S =(−12.00m)i^ + (14.00m)j^

A) W = 359.4J

B) W = -288.8J

C) W = -34.24J

Explanation:

From the question, we are given;

F = Constant horizontal Force = 30N

Angular direction of force = 37°

Now, since the angle at which this force is acting is counterclockwise from the +x-axis, the force vector will now be;

F' = (30Cos37)i^ + (30sin37)j^

F' = (23.96)i^ + (18.05)j^ - - - - eq(1)

We know that work done on a particle by constant force (F') during displacement is given by;

W = F'•s•cosθ

Now, θ is the angle between F' and s. However, from the work equation above, we can see that it has the form of the scalar product of 2 vectors i.e vector A x vector B = AB cosθ. Therefore, we can write work done as;

W = F'•s'

In light of this question, F' will represented as;

F' = (Fx)i^ + (Fy)j^ + (Fz)k^

Also,

s' = (x)i^ + (y)j^ + (z)k^

Thus, imputing thus values for F' and s' in W = F'•s', we obtain ;

W = (Fx)i^ + (Fy)j^ + (Fz)k^ • (x)i^ + (y)j^ + (z)k^

W = x(Fx) + y(Fy) + z(Fz) - - - - eq(2)

A) Displacement is given here as s' = (15.00m)i^

So, from s' equation, our displacement will be, x = 15m ; y= 0 ;z=0

Also, from equation 1,we see that;

Fx = 23.96 ; Fy = 18.05 ; Fz = 0

Thus,work done here, we follow equation 2;

W = (23.96 x 15) + (18.05 x 0) + (0x0)

W = 359.4J

B) Displacement is given here as s' = (-16.00m)j^

So, from s' equation, our displacement will be, x = 0m ; y=-16m ;z=0

Also, from equation 1,we see that;

Fx = 23.96 ; Fy = 18.05 ; Fz = 0

Thus,work done here, we follow equation 2;

W = (23.96 x 0) + (18.05 x -16) + (0x0)

W = -288.8J

C) Displacement is given here as s' = (−12.00m)i^ + (14.00m)j^

So, from s' equation, our displacement will be, x = - 12m ; y= 14m ;z=0

Also, from equation 1,we see that;

Fx = 23.96 ; Fy = 18.05 ; Fz = 0

Thus,work done here, we follow equation 2;

W = (23.96 x -12) + (18.05 x 14) + (0x0)

W = -34.24J

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

i don't know that i agree with that but what's the question ?