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A force \vec F = 5 \hat i - 4 \hat j +\hat k  moves a body of  mass 10 Kg from position (3,4,5) to a position (6,7,8) . Find the work (Joule) done 

Option: 1

8


Option: 2

10


Option: 3

6


Option: 4

4


Answers (1)

best_answer

As we have learned

Definition of work by constant force -

w = \underset{F}{\rightarrow}.\underset{S}{\rightarrow}

- wherein

The scalar product of the force vector (\underset{F}{\rightarrow}) and the displacement vector (\underset{S}{\rightarrow})

 

 

w = \underset{F}{\rightarrow}.\underset{S}{\rightarrow}

\vec {s_1}= 3 \hat i + 4 \hat j + 5 \hat k

\vec {s _2}= 6 \hat i + 7 \hat j + 8 \hat k

\vec{S}=\vec {s _2}- \vec {s _1}= 3 \hat i + 3 \hat j + 3 \hat k

 

w = \underset{F}{\rightarrow}.\underset{S}{\rightarrow} = = ( 5 \hat i - 4 \hat j + \hat k )\cdot ( 3 \hat i + 3 \hat j + 3 \hat k ) = 15- 12 +3 = 6 J

 

 

 

Posted by

Riya

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