(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.)

(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 .

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:

where the angle between a and bis θ  

(a)

Given that,

and

The work on the object is

=

={(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.

and

Let,

The power is -8.70 w.

∴ -8.70 =

⇒ - 8.70 = -2.7×a

⇒ a = 3.22

The velocity of the object is

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.