An object with mass 5 kg is under constant force (10i+20j)N

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An object with a mass of 5 kg is subjected to a constant force of (10i + 20j) N. The equation F = ma was utilized, leading to the formulation (10i + 20j) N = 5(vf - (-5i + 30j))/10. Through simplification and vector addition, the final velocity vector was determined to be (15i + 70j). This confirms the calculations were executed correctly, resulting in the accurate final velocity.

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vipson231
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An object with mass 5 kg is under constant force (10i+20j)N...

Homework Statement



(Question is in the attachment)

For the question I used the equation F=ma= m Δ(Vf-vi)/t and plugged in 5 for m and for
velocity I put (vf -(-5+30)/10= (10i + 20j)N. My final equation was

(10i+20j)N= 5(vf-(-5+30)/10

This is as far as I got and I can't solve the question =(. Am i doing something wrong?
 

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vipson231 said:
(10i+20j)N= 5(vf-(-5i+30j)/10
This is as far as I got and I can't solve the question =(. Am i doing something wrong?

Vector [itex]\equiv[/itex] fun with a direction. :-p
(10i + 20j) = 5([itex]\vec{v}[/itex][itex]_{f}[/itex]-(-5i + 30j))/10

Simplify...
(10i + 20j) = (1/2)([itex]\vec{v}[/itex][itex]_{f}[/itex]-(-5i + 30j))

Multiply vectors by the scalar
(10i + 20j) = (0.5)[itex]\vec{v}[/itex][itex]_{f}[/itex]-(-[itex]\frac{5}{2}[/itex]i + 15j)

Adding the "scaled" initial velocity vector to the other side
(10i + 20j) + (-[itex]\frac{5}{2}[/itex]i + 15j) = (0.5)[itex]\vec{v}[/itex][itex]_{f}[/itex]

Add vectors by components
(7.5i + 35j) = (0.5)[itex]\vec{v}[/itex][itex]_{f}[/itex]

(15i + 70j) = [itex]\vec{v}[/itex][itex]_{f}[/itex]

That's the third answer I see on the attachment, so I might have done this right.
 


Thanks a lot man! :D
 

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