Pushing a lawn mower, total force exerted?

[tbaskerv]

Homework Statement

A person pushes a 13.0 kg lawn mower at constant speed with a force of 77.0 N directed along the handle, which is at an angle of θ = 49.0° to the horizontal.

Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.8 m/s in 2.0 seconds with a retarding force of -50.5 N in the x direction.

Homework Equations

Using LaTeX Code: v^2 = v_0^2 + 2 a \\Delta x I found a = 0.81 m/s^2

The Attempt at a Solution

Using the acceleration of 0.81 m/s^2 times 13 kg I get F = 10.5 N, now I think I'm supposed to add the retarding force, but 61 N isn't correct.
Did I miss something?
Found the retarding force by finding the Fpcos(49) = 50.5 (in the x direction) Fp = 77.0 N.
Beyond that I'm lost. Please help

 

[tbaskerv]

Nevermind, another problem helped me out, thank you! More will come!

 

[nlightner]

I would like to clarify a few things about the given scenario. Firstly, the total force exerted on the lawn mower can be calculated by adding the force applied by the person (77.0 N) and the retarding force (-50.5 N), which gives a total force of 26.5 N. This is because the retarding force is acting in the opposite direction of the person's applied force.

Furthermore, the given information does not specify the direction and magnitude of the acceleration. In order to accurately calculate the force required to accelerate the lawn mower, we need to know the direction and magnitude of the acceleration. Additionally, the given acceleration value of 0.81 m/s^2 does not match with the given information of the lawn mower moving at a constant speed.

In order to accurately calculate the force required to accelerate the lawn mower, we need to know the following information:

1. The direction and magnitude of the acceleration.
2. The initial velocity of the lawn mower.
3. The final velocity of the lawn mower.

Once we have this information, we can use the equation F = ma to calculate the force required to accelerate the lawn mower.