Answer :
Answer:
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
[tex]\overrightarrow{\rm F}_x[/tex] = 30×cos 37° = 23.96 N
[tex]\overrightarrow{\rm F}_y[/tex] = 30×sin 37° = 18.0545 N
Therefore, [tex]\overrightarrow{\rm F}[/tex] = (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
[tex]\overrightarrow{\rm A}\cdot \overrightarrow{\rm B} = A\times B \times \cos \theta[/tex]
Therefore W = [tex]\overrightarrow{\rm F}\cdot \overrightarrow{\rm s}[/tex] where
[tex]\overrightarrow{\rm F}[/tex] = [tex]F_xi +F_yj + F_zk[/tex] and
[tex]\overrightarrow{\rm s}[/tex] = [tex]xi +yj + zk[/tex]
So that W = [tex](F_xi +F_yj + F_zk)\cdot (xi +yj + zk)[/tex]
= [tex]xF_xi +yF_yj + zF_zk[/tex]
So we look for the dot product
(a) When s = 15 m i
[tex]\overrightarrow{\rm s}[/tex] = [tex]xi +yj + zk[/tex] = [tex]15\cdot i +0\cdot j + 0\cdot k[/tex] and
W = [tex]xF_xi +yF_yj + zF_zk[/tex] = [tex]15 \times 23.96 N +0\times 18.0545N + 0\times 0N[/tex]
= 359.4 N·m = 359.4 J
(b) When s = -16 m j
[tex]\overrightarrow{\rm s}[/tex] = [tex]xi +yj + zk[/tex] = [tex]0\cdot i +15\cdot j + 0\cdot k[/tex] and
W = [tex]xF_xi +yF_yj + zF_zk[/tex] = [tex]0 \times 23.96 N +-16\times 18.0545N + 0\times 0N[/tex]
= -288.872 Nm = -288.872 J
(c) When s = -12.00 m i + 14.00 j, we have
W = [tex]-12 \times 23.96 N +14\times 18.0545N + 0\times 0N[/tex] = -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^
Answer:
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