Force Applied to a Box on a Flat Surface, Find the Force

[logan3]

Homework Statement

A box of mass 72 kg is at rest on a horizontal frictionless surface. A constant horizontal force of magnitude F then acts on the box, accelerating it to the right. You observe that it takes the box 3.4 seconds to travel 13 meters. What is the magnitude of the force F?

m = 72kg
t = 3.4s
s = 13m

Homework Equations

v = s/t
a = v/t
F = ma

The Attempt at a Solution

My first attempt:
v = s/t = (13m) / (3.4s) = 3.8235m/s
a = v/t = (3.8235m/s) / (3.4s) = 1.1245m/s^2
F = ma = (72kg)(1.1245m/s^2) = 80.968N == 81N

My second attempt:
v_i = 0m/s
a = F/m
s = (v_i)t + 0.5 (F/m)t^2
=> F = 2sm / t^2 = (2(13m)(72kg)) / (3.4s)^2 = 161.93N == 162N

Which answer is correct and why is one method correct and the other not?

Thank-you

 

[Satvik Pandey]

Hi logan.
Your second attempt looks good to me.I am not sure about calculations.
We use $v=s/t$ when there is uniform motion.But in this case there is uniform accelerated motion.