# Newton's Problems

1. Mar 30, 2010

### Paymemoney

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
F=ma

3. 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 $$\sqrt{39.62^2+22.22^2}$$
direction Resultant = $$tan^{-1}(\frac{22.22}{39.62})$$

direction = 30 degrees

find the magnitude $$F = m * \frac{v}{t}$$

$$F = 1.25 * \frac{45.43}{3.41}$$
F=16.65N

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

P.S

2. Mar 30, 2010

### rl.bhat

Horizontal = 17.4 + 26.8cos(34) = 39.62 m/s
You have to find the change in horizontal component of velocity (vf - vi)

3. Mar 30, 2010

### Paymemoney

is this the same in the vertical component?

edit never mind its not the same

Last edited: Mar 30, 2010
4. Mar 30, 2010

### rl.bhat

Yes. But initial vertical component of the velocity is zero.

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