(a) 34.6 N
To solve the problem, we have to analyze the forces acting along the horizontal direction.
We have:
- Forward: the component of the pull parallel to the ground, which is

where
F is the magnitude of the pull
is the angle
- Backward: the force of friction, which is

So, the equation of motion is

where
m = 20 kg is the mass of the wagon
a is the acceleration
In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

Therefore, we can find the pulling force:

(b) 43.9 N
In this case, the acceleration is

So, the equation of motion in this case is

So this time we have to take into account the term (ma).
Using the same data as before:
m = 20 kg


We find the new magnitude of F:
