12 kg wagon is being pulled at an angle of 38o 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? Is this question a typo? Am I supposed to solve for force or work?

Question

Physics - silverback, Tuesday, January 12, 2016 at 10:12pm
I need help
Physics - Damon, Tuesday, January 12, 2016 at 10:29pm
F cos 38 = m a

easy way is for constant acceleration
v average = (Vi+Vf)/2
so
v average = (0 + 2.2)/2 = 1.1 m/s
so time = t = 3.4/1.1 = 3.09 s
a = change in v / time
= 2.2 / 3.09
= .712 m/s^2
so
F cos 38 = 12 * .712

F = 10.8 Newtons
Physics - silverback, Tuesday, January 12, 2016 at 10:39pm
This is from one of your previous answers. You can do it with work if you wish or you can do it with F = m a

with work

work done = Force in direction of motion * distance
Force in direction of motion = F cos 38
= .788 F
work done = .788 F * 3.4 meters
= 2.68 F Joules

that is the increase in kinetic energy
(1/2) m v^2 = .5*12 * 2.2^2
= 29 Joules
so
2.68 F = 29
F = 10.8 Newtons
Why did you use kinetic energy is equal to work?

Answer 1

Conservation of energy
work done = increase of energy
or
work done = increase in kinetic energy + increase on potential energy

Answer 2

think lifting a barbell

work in = Force up * distance up
That is also the increase on potential energy m g h
when you drop it
that same value in Joules becomes (1/2) m v^2