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A particle of m=5kg is momentarily at re...

A particle of `m=5kg` is momentarily at rest at time `t=0`. It is acted upon by two forces `vec(F)_(1)` and `vec(F)_(2)`. `vec(F)_(1)`= `70hat(j)` N. The direction and manitude of `vec(F)_(2)` are unknown. The particle experiences a constant acceleration, `vec(a)`,in the direction as shown in figure. Neglect gravity.


a.Find the missing force `vec(F)_(2)`.
b. What is the velocity vector of the particle at `t=10 s`?
c. What third force, `vec(F)_(3)` is required to make the acceleration of the particle zero? Either give magnitude and direction of `vec(F)_(3)` or its components.

A

`30hati+40hatj`

B

`-(30hati+40hatj)`

C

`40hati+30hatj`

D

`-(40hati+30hatj)`

Text Solution

Verified by Experts

The correct Answer is:
B
The sume of forces `vecF_1+vecF_2=mveca`
`=5(10costhetahati+10sinthetahatj)=30hati+40hatj`
`vecF_1+vecF_2+vecF_3=0`
`vecF_3=-(vecF_1+vecF_2)=-30hati-40hatj`
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