InterviewSolution
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InterviewSolution
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The potential energy of a particle of mass 5 kg moving in xy-plane is given as `U=(7x + 24y)` joule, x and y being in metre. Initially at `t=0`, the particle is at the origin `(0,0)` moving with velovity of `(8.6hati+23.2hatj) ms^(1)`, ThenA. The velocity of the particle at `t=4s`, `5ms^(-1)`B. The acceleration of the particle is `5ms^(-2)`.C. The direction of motion of the particle initially (at t=0) is right angles to the direction of acceleration.D. The path of the particle is circle. |
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Answer» Correct Answer - A::B `F=-[(del)/(delX)hati + (delU)/(dely)hatj]=(7hati-24hatj) N` `a=F/m =(7/5hati - (24)/5hatj)m//s` `|a|=ssqrt((7/5)^(2) + (24/5)^(2) =5 m//s^(2))` Since, `a="constant"`, we can apply, `v=u+at` `=(8.6hati + 23.2hatj) + (-7/5hati-(24)/5hatj)(4)` `=(3hati +4hatj)m//s` `|v|=sqrt((3)^(2) + (4)^(2)=5m//s)` |
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