Answer :
Answer:
(a) The force which works on the object at a rate 15.89 w.
(b) The velocity of the object when the power is -8.7 w is [tex](3.22\ m/s)\hat j[/tex].
Explanation:
Work: The work on an object is the dot product of force that acts on the object and velocity of the object .
P = F.V
Dot product:
[tex]\vec a=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}[/tex]
[tex]\vec b=b_x\hat{i}+b_y\hat{j}+b_z\hat{k}[/tex]
[tex]\vec a.\vec b=(a_x\hat{i}+a_y\hat{j}+a_z\hat{k}).(b_x\hat{i}+b_y\hat{j}+b_z\hat{k})[/tex]
[tex]=a_x.b_x+a_y.b_y+a_z.b_z[/tex]
[tex]\vec a.\vec b=|a||b|cos\theta[/tex]
where the angle between a and bis θ
(a)
Given that,
[tex]\vec F=(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k[/tex]
and
[tex]\vec V=-(2.3 m/s) \hat i+(5.8 m/s)\hat j[/tex]
[tex]\Rightarrow \vec V=-(2.3 m/s) \hat i+(0m/s)\hat j+(5.8 m/s)\hat j[/tex]
The work on the object is
=[tex]\vec F.\vec V[/tex]
[tex]=\{(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k\}.\{-(2.3 m/s) \hat i+(0m/s)\hat j+(5.8 m/s)\hat j\}[/tex]
={(5.7)×(-2.3)+(-2.7)×0+(5.0×5.8)} w
=15.89 w
The instantaneous rate at which the force does work on the object is 15.89 w.
(b)
The velocity of the object consists of only a y component i.e the x component and y component are zero.
[tex]V_x=0[/tex] and [tex]V_z=0[/tex]
Let,
[tex]\vec V= 0\hat i+ a \hat j+0 \hat k[/tex]
The power is -8.70 w.
[tex]\vec F=(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k[/tex]
∴ -8.70 = [tex]\{(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k\}.( 0\hat i+ a \hat j+0 \hat k)[/tex]
⇒ - 8.70 = -2.7×a
[tex]\Rightarrow a=\frac{-8.7}{-2.7}[/tex]
⇒ a = 3.22
The velocity of the object is [tex](3.22\ m/s)\hat j[/tex]