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Question
A uniform force of (3^i+^j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2^i+^k) m to position (4^i+3^j−^k) m. The work done by the force on the particle is -
A uniform force of (3^i+^j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2^i+^k) m to position (4^i+3^j−^k) m. The work done by the force on the particle is -
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Solution
The correct option is D 9 J
The position vectors of initial and final position are,
→ri=2^i+^k, →rf=4^i+3^j−^k
The displacement vector is given by,
→S=→rf−→ri
∴→S=2^i+3^j−2^k
And, →F=3^i+^j
The work done is given by,
W=→F⋅→S
=3×2+1×3+0×(−2)=9 J
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
Hence, (B) is the correct answer.
The position vectors of initial and final position are,
→ri=2^i+^k, →rf=4^i+3^j−^k
The displacement vector is given by,
→S=→rf−→ri
∴→S=2^i+3^j−2^k
And, →F=3^i+^j
The work done is given by,
W=→F⋅→S
=3×2+1×3+0×(−2)=9 J
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.
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