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Newton's Problems

  • Thread starter Paymemoney
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  • #1
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Homework Statement


A 1.25kg object is moving in the x direction at 17.4m/s. Just 3.41s later, it is moving at 26.8m/s at 34.0 degrees to the x-axis. What are the magnitude and direction of the force applied during this time?


Homework Equations


F=ma

The Attempt at a Solution


Firstly i drew a diagram and it was in vector components so i calculated the resultant.

1st component
H: 17.4m/s
V: 0m/s

2nd component
H: 26.8cos(34)
V: 26.8sin(34)

Resultant component:
Horizontal = 17.4 + 26.8cos(34) = 39.62 m/s
Vertical = 26.8sin(34)

so [tex]\sqrt{39.62^2+22.22^2}[/tex]
direction Resultant = [tex]tan^{-1}(\frac{22.22}{39.62})[/tex]

direction = 30 degrees

find the magnitude [tex]F = m * \frac{v}{t}[/tex]

[tex]F = 1.25 * \frac{45.43}{3.41}[/tex]
F=16.65N

This is incorrect can someone tell me where i have gone wrong.

P.S
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
Horizontal = 17.4 + 26.8cos(34) = 39.62 m/s
You have to find the change in horizontal component of velocity (vf - vi)
 
  • #3
175
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is this the same in the vertical component?

edit never mind its not the same
 
Last edited:
  • #4
rl.bhat
Homework Helper
4,433
5
is this the same in the vertical component?
Yes. But initial vertical component of the velocity is zero.
 

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