A body is moving on a frictionless horizontal surface under the influence of a constant force F→$$=3i^+4j^$$ $$m/sec2$$. When the velocity of the body is $$2i^$$−$$3j^$$ m its kinetic energy.

Correct option is 3decreasing at a rate of 6 Joule / sec.

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A body is moving on a frictionless horizontal surface under the influence of a constant force $\vec{F} = (3 \hat{i} + 4 \hat{j}) m / \sec^{2} .$ When the velocity of the body is $(2 \hat{i} - 3 \hat{j}) m$ its kinetic energy.

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is increasing at a rate of 4 Joule / sec.

increasing at a rate of 5 Joule/sec.

decreasing at a rate of 6 Joule / sec.

decreasing at a rate of 4 Joule / sec.

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detailed solution

Correct option is C

(3)when v→=2i^−3j^ m/s, rate of work done on the body F→.v→=3i^+4j^. 2i^−3j^J/s=6−12 J/S=−6J/S.So by work - energy Theorem, kinetic energy is decreasing at a rate of 6 Joule / sec.

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A body is moving on a frictionless horizontal surface under the influence of a constant force F→=3i^+4j^ m/sec2. When the velocity of the body is 2i^−3j^ m its kinetic energy.

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A body is moving on a frictionless horizontal surface under the influence of a constant force $\vec{F} = (3 \hat{i} + 4 \hat{j}) m / \sec^{2} .$ When the velocity of the body is $(2 \hat{i} - 3 \hat{j}) m$ its kinetic energy.