A force F - - 4- - 6- - 3k- acting on a particle produces a displacement of S - - 2- - 3- - 5k- where F is expressed in Newtonn and S in the metre- Find the work done by the force-

Work = Dot product of F and S= F⃗ #183;S⃗ = ( 4î + 6ĵ+ 3k̂ ) #183; ( 2î + 3ĵ+ 5k̂ )= 8 + 18 + 15= 41 JouleWork done by the force is 41 Joule ...

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Vectors and Its Types

A force F →...

Question

A force $\to F = 4^i + 6^j + 3^k$ acting on a particle produces a displacement of $\to S = 2^i + 3^j + 5^k$ where F is expressed in Newtonn and S in the metre. Find the work done by the force.

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\to F = 3^i + c^j + 2^k

\to S = - 4^i + 2^j - 3^k

6 J

c

_{1} = (2^i + 3^j) m

_{2} = (4^i + 6^j) m

F = (3 x^{2}^i + 2 y^j) N

\to F = 3^i + c^j + 2^k

\to S = - 4^i + 2^j - 3^k

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(3^i +^j)

2 k g

(2^i +^k)

(4^i + 3^j -^k)

F = (3^i + c^j + 2^k) N

\to S = (- 4^i + 2^j - 3^k) m

6 J

^{'} c^{'}

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