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Laws of Motion question

2008 · Q152
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Laws of Motion question

2008 · Q152

NEETPhysicsLaws of MotionMCQ+4 / −1
Sand is being dropped on a conveyer belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of υ\upsilonυ m/s will be :
  1. A
    Mv2{{Mv} \over 2}2Mv​ newton
  2. B
    zero
  3. C
    Mvvv newton
  4. D
    2 Mvvv newton
View written solutionFree

Correct answer: C

F=ddt(Mv)=vdMdt+MdvdtF = {d \over {dt}}\left( {Mv} \right) = v{{dM} \over {dt}} + M{{dv} \over {dt}}F=dtd​(Mv)=vdtdM​+Mdtdv​

As v is a constant,

F=vdMdtF = v{{dM} \over {dt}}F=vdtdM​    

But dMdt=M kg/s{{dM} \over {dt}} = M\,kg/sdtdM​=Mkg/s

∴\therefore∴ To keep the conveyer belt moving at v m/s, force needed = vM newton.

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