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Q.

A body of mass 5 kg under the action of constant force F→=Fxi^+Fyj^ . It has velocity at t=0  as v→=(6i^-2j^)m/s and at t=10 s as v→=+6j^m/s· The force F→ is

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a

(-3j^+4j^)N

b

-35i^+45j^N

c

(3i^-4j^)N

d

35i^-45j^N

answer is A.

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Detailed Solution

Acceleration, a=v−ut=6j^−(6i^−2j^)10=−3i^+4j^5 m/s2 Force, F=ma=5×(−3i^+4j^)5=(−3i^+4j^)N
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