Work and energy problem feedback

[gbaby370]

I'm just starting to cover Work and Energy.

I had a question which asked, "A 12 kg wagon is being pulled at an angle of 38° above horizontal. What force is applied to the wagon if it accelerates from rest to a speed of 2.2 m/s over a distance of 3.4 m?"

For the record this is just a practice question and I am simply posting for feedback if I have made an error somewhere.

Given we have m= 12kg F=? Angle= 38 degrees V1=0 V2=2.2 Δd=3.4

I used the equation;

Work = Change in energy

W=ΔE

FcosθΔd=(0.5)mv2^2-(0.5)mv1^2

I rearranged and got.

F=(0.5)mv2^2/cosθΔd

F=(0.5)(12)(2.2^2)/cos38(3.4)

F=10.8N


How's that look?

The question might seem simple but I've been out of high school for several year and am hoping to go back to school, so I am just trying to ensure I'm remembering all of the fundamentals.

 

[gbaby370]

This basically what I did to rearrange the equation

 

[HallsofIvy]

I don't understand why you are asking about energy when the problem asks about force.

Force equals mass times acceleration so you can easily get the horizontal component of the applied component- the component that causes the acceleration. Then you need to use a little trig to get the applied force- cos(theta) is the horizontal force divided by the applied force- the hypotenuse in your force diagram.

 

[gbaby370]

This question was in the kinetic energy section.

But I think your method, and mine would equal the same thing?