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Five forces vec(F)(1) ,vec(F)(2) , vec(F...

Five forces `vec(F)_(1) ,vec(F)_(2) , vec(F)_(3) , vec(F)_(4)` , and `vec(F)_(5)` , are acting on a particle of mass `2.0kg` so that is moving with `4m//s^(2)` in east direction. If `vec(F)_(1)` force is removed, then the acceleration becomes `7m//s^(2)` in north, then the acceleration of the block if only `vec(F)_(1)` is action will be:

A

`16 ms^(-2)`

B

`sqrt65 ms^(-2)`

C

`sqrt260 ms^(-2)`

D

`sqrt33 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`vecF_1+vecF_2 +vecF_3+vecF_4+vecF_5=2(4hati)=8hati`
`vecF_2+vecF_3+vecF_4+vecF_5=2(7hatj)`
`= 14 hatj`
`vecF_1+14hatj=8hati`
`vecF_1=8hati-14hatj`
`F_1=sqrt((8)^2+(14)^2)=sqrt(260)=2sqrt65`
`F_1/m=(2sqrt65)/(2) = sqrt65 ms^(-2)`
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A moving particle of mass m is acted upon by five forces vec(F)_(1),vec(F)_(2),vec_(F)_(3),vec(F)_(4) and vec(F)_(5 . Forces vec(F)_(2) and vec(F)_(3) are conservative and their potential energy functions are U and W respectively. Speed of the particle changes from V_(a) to V_b when it moves from position a to b. Which of the following statement is/are true – (a) Sum of work done by vec(F)_(1),vec(F)_(4) and vec(F)_(5)=U_(b)-U_(a)+W_(b)-W_(a) (b) Sum of work done by vec(F)_(1),vec(F)_(4) and vec(F)_(5)=U_(b)-U_(a)+W_(b)-W_(a)+1/2m(V_(a)^(2)-V_(a)^(2)) (c) Sum of work done by all five forces =1/2m(V_(b)^(2)-V_(a)^(2)) (d) Sum of work done by vec(F)_(2) and vec(F)_(3)=(U_(b)+W_(b))-(U_(a)+W_(a))

Knowledge Check

  • When force vec(F)_(1),vec(F)_(2),vec(F)_(3) are acting on a particle of mass m , the particle remains in equilibrium. If the force vec(F)_(1) is now removed then the acceleration of the particle is :

    A
    `vec(F)_(1)//m`
    B
    `vec(F)_(1)//m`
    C
    `vec(F)_(2)-vec(F)_(3)//m`
    D
    `vec(F)_(2)//m`

  • When forces F_1 , F_2 , F_3 , are acting on a particle of mass m such that F_2 and F_3 are mutually perpendicular, then the particle remains stationary. If the force F_1 is now removed then the acceleration of the particle is

    A
    (a) `F//m`
    B
    (b) `F_2F_3//mF_1`
    C
    (c) `(F_2-F_3)//m`
    D
    (d) `F_2//m`

  • When force vec(F_(1)), vec(F_(2)),vec(F_(3))"…..."vec(F_(n)) act on a particle , the particle remains in equilibrium . If vec( F_(1)) is now removed then acceleration of the particle is

    A
    `(vec(F_(1)))/(m)`
    B
    `- (vec(F_(1)))/(m)`
    C
    `- ( vec( F_(2))+vec(F_(3))+"....."vec(F_(n))-vec(F_(1)))/(m)`
    D
    `(vec(F_(1)))/(m)`

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