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Question
A particle is shifted from point (0,0,1 m) to (1 m,1 m,2 m), under the simultaneous action of several forces. Two of the forces are −→F1=(2^i+3^j−^k) N and −→F2=(^i−2^j+2^k) N. Find the work done by these two forces
A particle is shifted from point (0,0,1 m) to (1 m,1 m,2 m), under the simultaneous action of several forces. Two of the forces are −→F1=(2^i+3^j−^k) N and −→F2=(^i−2^j+2^k) N. Find the work done by these two forces
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Solution
The correct option is B 5 J
As particle is shifted from point (0,0,1 m) to (1 m,1 m,2 m)
Displacement vector Δ→r=(^i+^j+^k)
net force acting is −→F1+−→F2=(2^i+3^j−^k) N+(^i−2^j+2^k) N=(3^i+^j+^k)
Work done by a constant force equals to the dot product of the force and displacement vectors.
W=→F.Δ→r⇒W=(−→F1+−→F2).Δ→r
Substituting the given values, we have W=(3^i+^j+^k).(^i+^j+^k)=3+1+1=5 J
As particle is shifted from point (0,0,1 m) to (1 m,1 m,2 m)
Displacement vector Δ→r=(^i+^j+^k)
net force acting is −→F1+−→F2=(2^i+3^j−^k) N+(^i−2^j+2^k) N=(3^i+^j+^k)
Work done by a constant force equals to the dot product of the force and displacement vectors.
W=→F.Δ→r⇒W=(−→F1+−→F2).Δ→r
Substituting the given values, we have W=(3^i+^j+^k).(^i+^j+^k)=3+1+1=5 J
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