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

(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`.

Step-by-step 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=(-8.7)/(-2.7)`

⇒ a = 3.22

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