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A body is displaced from −→rA=(2^i+4^j−2^k) m to −→rB=(8^i+14^j+6^k) m under the action of a constant force →F=(2^i+3^j+6^k) N.If the initial kinetic energy of the body is 15 J, Find the kinetic energy of the body at the end of displacement. |
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Answer» A body is displaced from −→rA=(2^i+4^j−2^k) m to −→rB=(8^i+14^j+6^k) m under the action of a constant force →F=(2^i+3^j+6^k) N. |
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