Question

(a) At a certain instant, a particle-like object is acted on by a force F= (5.7 N)i -(2.7 N)j + (5.0 N)k while the object's velocity is v= -(2.3 m/s)i + (5.8 m/s)k. What is the instantaneous rate at which the force does work on the object?
(b) At some other time, the velocity consists of only a y component. If the force is unchanged, and the instantaneous power is -8.70 W, what is the velocity of the object just then? (Give your answer without a unit vector.)

Answer 1

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 $(3.22\ m/s)\hat j$.

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:

$\vec a=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}$

$\vec b=b_x\hat{i}+b_y\hat{j}+b_z\hat{k}$

$\vec a.\vec b=(a_x\hat{i}+a_y\hat{j}+a_z\hat{k}).(b_x\hat{i}+b_y\hat{j}+b_z\hat{k})$

     $=a_x.b_x+a_y.b_y+a_z.b_z$

$\vec a.\vec b=|a||b|cos\theta$

where the angle between a and bis θ  

(a)

Given that,

$\vec F=(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k$

and

$\vec V=-(2.3 m/s) \hat i+(5.8 m/s)\hat j$

$\Rightarrow \vec V=-(2.3 m/s) \hat i+(0m/s)\hat j+(5.8 m/s)\hat j$

The work on the object is

=$\vec F.\vec V$

$=\{(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k\}.\{-(2.3 m/s) \hat i+(0m/s)\hat j+(5.8 m/s)\hat j\}$

={(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.

$V_x=0$ and $V_z=0$

Let,

$\vec V= 0\hat i+ a \hat j+0 \hat k$

The power is -8.70 w.

$\vec F=(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k$

∴ -8.70 = $\{(5.7N)\hat i-(2.7N) \hat j+(5.0N) \hat k\}.( 0\hat i+ a \hat j+0 \hat k)$

⇒ - 8.70 = -2.7×a

$\Rightarrow a=\frac{-8.7}{-2.7}$

⇒ a = 3.22

The velocity of the object is $(3.22\ m/s)\hat j$