Question 6.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
F= -i + 2j + 3k N
Where are i,j,k are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Chapter Work, Energy And Power Page 136
Answers
Answered by
4
Hey there !!!!!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Force applied on body = -i+2j+3k
With a displacement of 4m along z axis
s= 4k
Work done = F.s = (-i+2j+3k)(4k) = 12J
So work done by body when it moves 4m along z axis is 12Joules
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope this helped you.............................
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Force applied on body = -i+2j+3k
With a displacement of 4m along z axis
s= 4k
Work done = F.s = (-i+2j+3k)(4k) = 12J
So work done by body when it moves 4m along z axis is 12Joules
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope this helped you.............................
Answered by
16
Given,
F =( -i + 2j + 3k ) N
distance of 4m along z axis .
e.g S = 4k m
we know,
work done is the dot product of Force and displacement vectors .
e.g W = F.S
= ( -i + 2j + 3k).(4k )
= (-i + 2j + 3k)( 0i + 0j + 4k)
= -1×0 + 2×0 + 3×4
= 12 Joule
hence, work done = 12 Joule
F =( -i + 2j + 3k ) N
distance of 4m along z axis .
e.g S = 4k m
we know,
work done is the dot product of Force and displacement vectors .
e.g W = F.S
= ( -i + 2j + 3k).(4k )
= (-i + 2j + 3k)( 0i + 0j + 4k)
= -1×0 + 2×0 + 3×4
= 12 Joule
hence, work done = 12 Joule
Similar questions