A. -20
B. -10
C. 13
D. 15
E. -30
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SOLUTION
The work done (W) in moving an object in a force field is given by the formula:
W = F × d
Where F is the force field and d is the differential displacement vector.
Given F = 4i – 3j + 5 k and the initial and final positions (1, -2, 2) and (-2, -1, 3), respectively, you can find d as the difference between the final and initial position vectors.
Let P represent the initial position and Q the final position of the object.
Therefore displacement is the vector between P and Q
P (1, -2, 3) and Q (-2, -1, 3)
d = Q – P
d =(-2, -1, 3) – (1, -2, 2)
d = -2-(-1) , -1-(-2) , 3-(2)
d= (-3, 1, 1).
d = -3i + j + k
Given Force = F = 4i – 3j + 5k
Work done
W= F × d(Q-P)
= (4i – 3j + 5k) × (-3i + j + k)
= -12 -3 + 5
W = -10
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Work done by force acting on an object/body can be described as the formula below
W= work done
F= force acting
s displacement of the object/body
So work is force multiplied by displacement
But when the force and displacement or given in vector form we use dot product or scalar product as the work done is a scalar
, like below
Here the force the force is already given as
And we are given that it moves from point (1,2,-3) to the point (2,0,-5)
The displacement vector s with initial point (x1, y1, z1) and terminal point (x2, y2, z2) is s = (x2 − x1, y2 − y1, z2 − z1).
So now work done is
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QUESTION 3: The work done in moving an object along a straight line from (3, 2, -1) to (2, -1, 4) in a force field given by, F = 4i – 3j + 2k, is?
SOLUTION
