Question

1. The constant force (3 î - 8 ĵ - 11 ƙ ) N is applied over a displacement (6j - 7ƙ) m. What is the work done by this force during this process? Answer in joules.

2. Whenever net work is done on an object, what will always be true?

the height of the object will change

the energy of the object will change

the gravitational potential energy of the object will change

the speed of the object will change

heat is produced

Answer 1

1.
Given Constant Force vector is

   F = (3 î - 8 ĵ - 11 ƙ ) N

applied over a displacement S = (6j - 7ƙ) m.

the work done is W = F . S

scalar product of the two vectors Force and displacement

   W = F.S

   W = (3 î - 8 ĵ - 11 ƙ ). (0 i + 6j - 7ƙ) J

   W = (3*0)(i.0) + ((-8)*6) (j.j)) + ((-11)(-7))(k.k)

we know that (i.i) = (j.j) = (k.k) = 1

so work done W = (0 -48*1+77*1)

       W = (29) J

the work done by the force is 29 J

2. if the network don on an object then
First option
always the height of the object always may not be change , for example if the object is on the surface then the force applied on it if the object makes displacement on the surface then work will also done on it , but the object will be on the surface, so it is not true

third option

gravitational potential energy of the object may not be changed always as explained in the case of first option

fourth option

Speed of the object may not change always, the object may move with uniform velocity

fifth option

heat is produced

it is not true always , if there exists dissipative forces there may be heat produced

so the option TWO will be the appropriate answer

that is the Energy of the object will change

that is if the object is at rest upon applying the force gets displaced and move with velocity so it would have kinetic energy, intially which is zero
and if the object raised to a height h, from ground (h=0 m ), then its gravitational potenial energy is u = mgh , then its energy will change

Answer is the energy of the object will change.